2.5 Limits at Infinity

For Use Only in 2013 2014– Pilot Program

96 Chapter 2 r Limits

46–47. Steep secant lines b. Create a graph that gives a more complete representation of f. a. Given the graph of f in the following figures, find the slope of the 1 2 1 1 22 y secant line that passes through 0, 0 and h, f h in terms of h, for h 7 0 and h 6 0. 20 b. Evaluate the limit of the slope of the secant line found in part (a) as h S 0+ and h S 0-. What does this tell you about the line tangent to the curve at 10, 02? 15 1 2 = 1>3 46. f x x 2000 10 y ϭ y 50 ϩ 100 x2

5 (h, h1/3) (0, 0) Ϫ4 Ϫ2 420 x h x y ϭ x1/3 Technology Exercises T 49–56. Asymptotes Use analytical methods and/or a graphing utility to identify the vertical asymptotes (if any) of the following functions.

x2 - 3x + 2 2>3 1 2 = 1 2 = - 2 1 2 = 47. f x x 49. f x 50. g x 2 ln x x10 - x9 y x p 1 2 = e 1 2 = a xb 6 y ϭ x2/3 51. h x 52. p x sec , for x 2 1x + 123 2 (h, h2/3) pu p 53. g1u2 = tan a b 54. q1s2 = (0, 0) 10 s - sin s h x 1 2 = 1 1 2 = 1>x 55. f x 56. g x e 1x sec x T 57. Can a graph intersect a vertical asymptote? A common mis- conception is that the graph of a function never intersects its vertical asymptotes. Let T 48. Care with graphing The figure shows the graph of the function 1 2 = 2000 3 - 4 * 3 4 4 6 f x graphed in the window 4, 4 0, 20 . if x 1 2 1 2 = - 50 + 100x f x W x 1 2 Ú . 1 2 1 2 1 2 x if x 1 a. Evaluate lim f x , lim f x , and lim f x . xS0+ xS0- xS0 Explain why x = 1 is a vertical asymptote of the graph of f and show that the graph of f intersects the line x = 1.

QUICK CHECK ANSWERS 1. Answers will vary, but all graphs should have a vertical asymptote at x = 2. 2. - ∞; ∞ 3. As x S -4+ , x 6 0 and 1 + 2 7 1 + 2 S x 4 0, so x x 4 0 through negative values. 1x - 121x - 22 = 1 - 2 = 4. lim lim x 1 1, which is not

xS2 x - 2 xS2 an infinite limit, so x = 2 is not a vertical asymptote. ➤

2.5 Limits at Infinity

Limits at infinity—as opposed to infinite limits—occur when the independent variable becomes large in magnitude. For this reason, limits at infinity determine what is called the end behavior of a function. An application of these limits is to determine whether a system (such as an ecosystem or a large oscillating structure) reaches a steady state as time increases.

2.5 Limits at Infinity 97

y lim f (x) q Limits at Infinity and Horizontal Asymptotes xᠬ 1 2 = -1 1- ∞ ∞2 q Consider the function f x tan x, whose domain is , (Figure 2.30). As S ∞ 1 2 p> x becomes arbitrarily large (denoted x ), f x approaches 2, and as x becomes Horizontal 1 S - ∞ 1 2 -p> f (x) tan x arbitrarily large in magnitude and negative (denoted x ), f x approaches 2. asymptote These limits are expressed as

- p - p 2 x lim tan 1 x = and lim tan 1 x = - . xS∞ 2 xS-∞ 2 Horizontal asymptote The graph of f approaches the horizontal line y = p>2 as x S ∞, and it approaches the q horizontal line y = -p>2 as x S - ∞. These lines are called horizontal asymptotes. lim f (x) q xᠬ FIGURE 2.30 DEFINITION Limits at Infinity and Horizontal Asymptotes 1 2 y If f x becomes arbitrarily close to a finite number L for all sufficiently large and posi- lim f (x) L tive x, then we write xᠬ L 1 2 = lim f x L. S f (x) y f (x) x ∞ 1 2 = x ៬ We say the limit of f x as x approaches infinity is L. In this case, the line y L is a horizontal asymptote of f (Figure 2.31). The limit at negative infinity, x x 1 2 = lim f x M, is defined analogously. When the limit exists, the horizontal ៬ xS - ∞ f (x) asymptote is y = M. M lim f (x) M

xᠬ x ➤ FIGURE 2.31 QUICK CHECK 1 Evaluate x>1x + 12 for x = 10, 100, and 1000. What is lim ? xS∞ x + 1

EXAMPLE 1 Limits at infinity Evaluate the following limits.

a + 10 b a + 3 sin x b a. lim 2 2 b. lim 3 ➤ The limit laws of Theorem 2.3 and the xS - ∞ x xS∞ 1x Squeeze Theorem apply if x S a is SOLUTION replaced with x S ∞ or x S - ∞. a. As x becomes large and negative, x2 becomes large and positive; in turn, 10>x2 y approaches 0. By the limit laws of Theorem 2.3,

10 10 10 f (x) 2 a + b = + a b = + = x2 lim 2 2 lim 2 lim 2 2 0 2. xS-∞ x xS-∞ (+xS+)++*-∞ x c equals 2 equals 0 lim f (x) 2 lim f (x) 2 10 xᠬ xᠬ Notice that lim a2 + b is also equal to 2. Therefore, the graph of y = 2 + 10>x2 xS∞ x2 = S ∞ S - ∞ 2 approaches the horizontal asymptote y 2 as x and as x (Figure 2.32). y 2 b. The numerator of sin x> 1x is bounded between -1 and 1; therefore, for x 7 0, x 2 1 sin x 1 - … … . FIGURE 2.32 1x 1x 1x As x S ∞, 1x becomes arbitrarily large, which means that -1 1 lim = lim = 0. xS∞ 1x xS∞ 1x sin x It follows by the Squeeze Theorem (Theorem 2.5) that lim = 0. xS∞ 1x

98 Chapter 2 r Limits y Using the limit laws of Theorem 2.3, 6 3 sin x sin x lim a3 + b = lim 3 + 3 lim a b = 3. 3 sin x S∞ 1 S∞ S∞ 1 5 f (x) 3 x x x (+x +)++*x x e equals 3 equals 0 4 3 sin x 3 The graph of y = 3 + approaches the horizontal asymptote y = 3 as x 1x 2

lim f (x) 3 becomes large (Figure 2.33). Note that the curve intersects its asymptote infinitely x ➤ 1 ៬ many times. Related Exercises 9–14

0 0 1020 30 40 x Infinite Limits at Infinity FIGURE 2.33 It is possible for a limit to be both an infinite limit and a limit at infinity. This type of limit 1 2 occurs if f x becomes arbitrarily large in magnitude as x becomes arbitrarily large in lim f (x) magnitude. Such a limit is called an infinite limit at infinity and is illustrated by the func- xᠬ y 1 2 = 3 tion f x x (Figure 2.34).

DEFINITION Infinite Limits at Infinity 1 1 2 f (x) x3 If f x becomes arbitrarily large as x becomes arbitrarily large, then we write 1 2 = ∞ lim f x . xS∞ 1 1 x 1 2 = - ∞ 1 2 = ∞ 1 2 = - ∞ The limits lim f x , lim f x , and lim f x are S∞ S - ∞ S - ∞ x x x 1 defined similarly.

Infinite limits at infinity tell us about the behavior of polynomials for large-magnitude lim f (x) 1 2 = n values of x. First, consider power functions f x x , where n is a positive integer. xᠬ Figure 2.35 shows that when n is even, lim xn = ∞, and when n is odd, lim xn = ∞ S { ∞ S∞ FIGURE 2.34 and lim xn = - ∞. x x xS - ∞

5 n 0 even: y n 0 odd: y y x n n n lim x lim x lim x y x7 x ៬ x ៬ x ៬ 60 y x6 y x4

20 y x3 40

3 2 1 2 3 x 20

20 y x2

3 2 1 1 2 3 x

FIGURE 2.35

2.5 Limits at Infinity 99

1 2 = > n = -n It follows that reciprocals of power functions f x 1 x x , where n is a positive integer, behave as follows:

1 = -n = 1 = -n = lim n lim x 0 and lim n lim x 0. xS∞ x xS∞ xS - ∞ x xS - ∞ S { ∞ QUICK CHECK 2 Describe the behavior From here, it is a short step to finding the behavior of any polynomial as x . Let 1 2 = n + n - 1 + g + 2 + +

of p1x2 = -3x3 as x S ∞ and as p x anx an - 1x a2x a1x a0. We now write p in the equivalent x S - ∞. ➤ form a - a - a 1 2 = n ° + n 1 + n 2 + g+ 0 ¢ p x x an 2 n . x x "x e e S0 S0 S0

Notice that as x becomes large in magnitude, all the terms in p except the first term ap- S {∞ 1 2 ≈ n proach zero. Therefore, as x , we see that p x anx . This means that as S { ∞ n x , the behavior of p is determined by the term anx with the highest power of x.

THEOREM 2.6 Limits at Infinity of Powers and Polynomials Let n be a positive integer and let p be the polynomial 1 2 = n + n - 1 + + 2 + + ≠ p x anx an - 1x g a2x a1x a0, where an 0. 1. lim xn = ∞ when n is even. xS { ∞ 2. lim xn = ∞ and lim xn = - ∞ when n is odd. xS∞ xS - ∞ 1 = -n = 3. lim n lim x 0. xS { ∞ x xS { ∞ 1 2 = n = ∞ - ∞ 4. lim p x lim an x or , depending on the degree of the xS { ∞ xS { ∞ polynomial and the sign of the leading coefficient an.

EXAMPLE 2 Limits at infinity Evaluate the limits as x S {∞ of the following functions. a. p1x2 = 3x4 - 6x2 + x - 10 b. q1x2 = -2x3 + 3x2 - 12 SOLUTION a. We use the fact that the limit is determined by the behavior of the leading term: 1 4 - 2 + - 2 = 4 = ∞ lim 3x 6x x 10 lim 3 x . S S b x ∞ x ∞ S ∞ Similarly, lim 13x4 - 6x2 + x - 102 = lim 3 x4 = ∞.

S S b x - ∞ x - ∞ S ∞

b. Noting that the leading coefficient is negative, we have lim 1-2x3 + 3x2 - 122 = lim 1-2 x32 = - ∞

S∞ S∞ b x x S ∞ lim 1-2x3 + 3x2 - 122 = lim 1-2 x32 = ∞.

S S b

x - ∞ x - ∞ S -∞ Related Exercises 15–24 ➤

100 Chapter 2 r Limits

End Behavior The behavior of polynomials as x S {∞ is an example of what is often called end behavior. Having treated polynomials, we now turn to the end behavior of rational, algebraic, and transcendental functions.

EXAMPLE 3 End behavior of rational functions Determine the end behavior for the following rational functions. + 4 + 2 - 3 - + 1 2 = 3x 2 1 2 = 40x 4x 1 1 2 = x 2x 1 a. f x b. g x c. h x x2 - 1 10x4 + 8x2 + 1 2x + 4 SOLUTION a. An effective approach for evaluating limits of rational functions at infinity is to divide both the numerator and denominator by xn, where n is the largest power appearing in the denominator. This strategy forces the terms corresponding to lower powers of x to approach 0 in the limit. In this case, we divide by x2:

approaches 0 + 3x 2 3 + 2 3x + 2 x2 x x2 0 lim = lim = lim = = 0. xS∞ x2 - 1 xS∞ x2 - 1 xS∞ 1 1 1 - x2 x2 b approaches 0 3x + 2 ➤ Recall that the degree of a polynomial is A similar calculation gives lim = 0, and thus the graph of f has the horizontal S - ∞ 2 - the highest power of x that appears. x x 1 asymptote y = 0. You should confirm that the zeros of the denominator are -1 and 1, which correspond to vertical asymptotes (Figure 2.36). In this example, the y 3x 2 f (x) degree of the polynomial in the numerator is less than the degree of the polynomial in x2 1 the denominator. b. Again we divide both the numerator and denominator by the largest power appearing 1 in the denominator, which is x4: 4 2 1 1 x 40x + 4x - 1 lim f (x) 0 40x4 + 4x2 - 1 4 4 4 Divide the numerator and x x x x ᠬ lim = lim 4 lim f (x) 0 xS∞ 10x4 + 8x2 + 1 xS∞ 10x4 8x2 1 denominator by x . xᠬ + + x4 x4 x4 approaches 0 approaches 0 b b FIGURE 2.36 + 4 - 1 40 2 4 y x x = lim Simplify. S∞ lim g(x) 4 lim g(x) 4 x 8 1 10 + + xᠬ xᠬ x2 x4

4 b b approaches 0 approaches 0 40 + 0 + 0 2 = = 40x4 4x2 1 4. Evaluate limits. g(x) 10 + 0 + 0 10x4 8x2 1 Using the same steps (dividing each term by x4), it can be shown that x 4 2 2 4 40x4 + 4x2 - 1 lim = 4. This function has the horizontal asymptote y = 4 xS - ∞ 10x4 + 8x2 + 1 (Figure 2.37). Notice that the degree of the polynomial in the numerator equals the FIGURE 2.37 degree of the polynomial in the denominator.

2.5 Limits at Infinity 101

c. We divide the numerator and denominator by the largest power of x appearing in the denominator, which is x, and then take the limit: 3 x - 2x + 1 x3 - 2x + 1 x x x Divide numerator and lim = lim S∞ + S∞ denominator by x. x 2x 4 x 2x + 4 x x approaches 0 arbitrarily large constant b b b 1 x2 - 2 + x = lim Simplify. xS∞ 4 2 + b x constant b approaches 0 = ∞. Take limits. As x S ∞, all the terms in this function either approach zero or are constant—except the x2@term in the numerator, which becomes arbitrarily large. Therefore, the limit of the x3 - 2x + 1 function does not exist. Using a similar analysis, we find that lim = ∞. xS-∞ 2x + 4 These limits are not finite, and so the graph of the function has no horizontal asymptote. In this case, the degree of the polynomial in the numerator is greater than the degree of

the polynomial in the denominator. Related Exercises 25–34 ➤

The conclusions reached in Example 3 can be generalized for all rational functions. These results are summarized in Theorem 2.7 (Exercise 74).

THEOREM 2.7 End Behavior and Asymptotes of Rational Functions p1x2 1 2 = Suppose f x is a rational function, where q1x2 1 2 = m + m- 1 + + 2 + + p x amx am- 1x g a2x a1x a0 and 1 2 = n + n - 1 + + 2 + + q x bnx bn - 1x g b2x b1x b0, ≠ ≠ with am 0 and bn 0. a. Degree of numerator less than degree of denominator If m 6 n, then 1 2 = = lim f x 0, and y 0 is a horizontal asymptote of f . QUICK CHECK 3 Use Theorem 2.7 to xS { ∞ find the vertical and horizontal

b. Degree of numerator equals degree of denominator If m = n, then = 10x ➤ 1 2 = > = > asymptotes of y . lim f x am bn, and y am bn is a horizontal asymptote of f . 3x - 1 xS { ∞ c. Degree of numerator greater than degree of denominator If m 7 n, then 1 2 = ∞ - ∞ lim f x or , and f has no horizontal asymptote. xS { ∞ 1 2 d. Assuming that f x is in reduced form (p and q share no common factors), vertical asymptotes occur at the zeros of q.

102 Chapter 2 r Limits

Although it isn’t stated explicitly, Theorem 2.7 implies that a rational function can have at most one horizontal asymptote, and whenever there is a horizontal asymptote, p1x2 p1x2 lim = lim . The same cannot be said of other functions, as the next examples xS∞ q1x2 xS - ∞ q1x2 show.

EXAMPLE 4 End behavior of an algebraic function Examine the end behavior of 3 - 2 + 1 2 = 10x 3x 8 f x . 225x6 + x4 + 2 SOLUTION The square root in the denominator forces us to revise the strategy used with rational functions. First, consider the limit as x S ∞. The highest power of the polynomial in the denominator is 6. However, the polynomial is under a square root, so effectively, the highest power in the denominator is 2x6 = x3. Dividing the numerator and denominator by x3, for x 7 0, the limit is evaluated as follows: 3 2 10x - 3x + 8 10x3 - 3x2 + 8 x3 x3 x3 lim = lim Divide by 2x6 = x3. xS∞ 2 6 + 4 + xS∞ 6 4 25x x 2 25x + x + 2 B x6 x6 x6 approaches 0 approaches 0 b b 3 8 10 - + x x3 = lim Simplify. xS∞ 1 2 25 + + A x2 x6 b b approaches 0 approaches 0 10 = = 2. Evaluate limits. 125 ➤ Recall that As x S - ∞, x3 is negative, so we divide numerator and denominator by 2x6 = -x3 x if x Ú 0 2x 2 = x = b (which is positive): -x if x 6 0. 3 2 Therefore, 10x - 3x + 8 3 2 x3 if x Ú 0 10x - 3x + 8 -x3 -x3 -x3 Divide by 2x6 = x3 = b lim = lim - 3 6 . S S 2 6 = - 3 7 x if x 0 x - ∞ 225x6 + x4 + 2 x - ∞ 25x6 x4 2 x x 0. Because x is negative as x S - ∞, we have + + B 6 6 6 2x6 = -x3. x x x approaches 0 approaches 0 b b 3 8 -10 + - y x x3 = lim Simplify. xS - ∞ 1 2 6 25 + + A x2 x6 b b 10x3 3x2 8 approaches 0 approaches 0 f (x) ͙ 6 4 25x x 2 10 = - = -2. Evaluate limits. 125 1 lim f (x) 2 The limits reveal two asymptotes, y = 2 and y = -2. Observe that the graph crosses xᠬ

both horizontal asymptotes (Figure 2.38). x ➤ 21 Related Exercises 35–38

2 EXAMPLE 5 End behavior of transcendental functions Determine the end behav- lim f (x) 2 xᠬ ior of the following transcendental functions. 1 2 = x 1 2 = -x 1 2 = 1 2 = FIGURE 2.38 a. f x e and g x e b. h x ln x c. f x cos x

2.5 Limits at Infinity 103

y SOLUTION 1 2 = x S ∞ x lim ex a. The graph of f x e (Figure 2.39) makes it clear that as x , e increases with- xᠬ out bound. All exponential functions bx with b 7 1 behave this way, because raising a number greater than 1 to ever-larger powers produces numbers that increase without bound. The figure also suggests that as x S - ∞, the graph of ex approaches the hori- f (x) ex zontal asymptote y = 0. This claim is confirmed analytically by recognizing that

lim ex 0 x = -x = 1 = x lim e lim e lim x 0. ᠬ xS - ∞ xS∞ xS∞ e 0 x Therefore, lim ex = ∞ and lim ex = 0. Because e-x = 1>ex, it follows that xS∞ xS - ∞ lim e-x = 0 and lim e-x = ∞. FIGURE 2.39 xS∞ xS - ∞ x y f (x) e b. The domain of ln x is 5x: x 7 06, so we evaluate lim ln x and lim ln x to determine y x xS0 + xS∞ end behavior. For the first limit, recall that ln x is the inverse of ex (Figure 2.40), and the graph of ln x is a reflection of the graph of ex across the line y = x. The horizontal lim ln x asymptote 1y = 02 of ex is also reflected across y = x, becoming a vertical asymptote x ៬ 1x = 02 for ln x. These observations imply that lim ln x = - ∞. S + 1 x 0 h(x) f 1(x) ln x It is not obvious whether the graph of ln x approaches a horizontal asymptote or whether the function grows without bound as x S ∞. Furthermore, the numerical 1 x Reflection of y ex evidence (Table 2.9) is inconclusive because ln x increases very slowly. The inverse across line y x relation between ex and ln x is again useful. The fact that the domain of ex is 1- ∞,∞2 implies that the range of ln x is also 1- ∞,∞2. Therefore, the values of ln x lie in the lim ln x x 0 interval 1- ∞, ∞2, and it follows that lim ln x = ∞. ៬ xS∞ FIGURE 2.40 c. The cosine function oscillates between -1 and 1 as x approaches infinity (Figure 2.41). Therefore, lim cos x does not exist. For the same reason, lim cos x does not exist. xS∞ xS - ∞ Table 2.9 y x x ln f (x) cos x 1 10 2.302 105 11.513 10 10 23.026 1050 115.129 x 1099 227.956 TT 1 ∞ ??? lim cos x does not exist. lim cos x does not exist. x៬ x៬

FIGURE 2.41 Related Exercises 39–44 ➤

The end behavior of exponential and logarithmic functions are important in upcoming work. We summarize these results in the following theorem.

x -x THEOREM 2.8 End Behavior of e , e , and ln x The end behavior for ex and e-x on 1- ∞, ∞2 and ln x on 10, ∞2 is given by the QUICK CHECK 4 How do the functions following limits:

e10x and e-10x behave as x S ∞ and as ➤ x = ∞ x = x S - ∞? lim e and lim e 0, xS∞ xS - ∞ -x = -x = ∞ lim e 0 and lim e , xS∞ xS - ∞ lim ln x = - ∞ and lim ln x = ∞. xS0 + xS∞

104 Chapter 2 r Limits

SECTION 2.5 EXERCISES 1 2 1 2 Review Questions 35–38. Algebraic functions Evaluate lim f x and lim f x for the 1 2 = xS∞ xS - ∞ 1. Explain the meaning of lim f x 10. xS - ∞ following functions. Then give the horizontal asymptote(s) of f (if any). 2. What is a horizontal asymptote? 3 + 1 2 = 4x 1 1 2 f x 35. f x 1 2 S 1 2 S ∞ S ∞ 3 + 2 6 + 3. Determine lim if f x 100,000 and g x as x . 2x 16x 1 xS∞ g1x2 2x 2 + 1 1 2 = -2x 1 2 = 4. Describe the end behavior of g x e . 36. f x 2x + 1 1 2 = - 3 5. Describe the end behavior of f x 2x . 23 6 + 1 2 = x 8 37. f x 6. The text describes three cases that arise when examining the end 2 4 1 2 = 1 2> 1 2 4x + 23x + 1 behavior of a rational function f x p x q x . Describe the 1 2 = 1 - 2 2 + 2 end behavior associated with each case. 38. f x 4x 3x 9x 1 x x -x 7. Evaluate lim e , lim e , and lim e . 39–44. Transcendental functions Determine the end behavior of the xS∞ xS - ∞ xS∞ 1 2 = following transcendental functions by evaluating appropriate limits. 8. Use a sketch to find the end behavior of f x ln x. Then provide a simple sketch of the associated graph, showing asymp- Basic Skills totes if they exist. 9–14. Limits at infinity Evaluate the following limits. 1 2 = - -x 1 2 = x 1 2 = - 39. f x 3e 40. f x 2 41. f x 1 ln x 10 1 10 9. lim a3 + b 10. lim a5 + + b 1 2 = 1 2 = 1 2 = 50 S 2 S 2 42. f x ln x 43. f x sin x 44. f x x ∞ x x ∞ x x e2x cos u 3 + 2x + 4x2 Further Explorations 11. lim 12. lim uS∞ u2 xS∞ x2 45. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. cos x5 100 sin4 x3 13. lim 14. lim a5 + + b a. The graph of a function can never cross one of its horizontal xS∞ 1x xS - ∞ x x2 asymptotes. 1 2 = 15–24. Infinite limits at infinity Determine the following limits. b. A rational function f can have both lim f x L and xS∞ 1 2 = ∞ 12 11 lim f x . 15. lim x 16. lim 3 x S - ∞ xS∞ xS-∞ x c. The graph of any function can have at most two horizontal - - 17. lim x 6 18. lim x 11 asymptotes. xS∞ xS-∞ 19. lim 13x12 - 9x72 20. lim 13x7 + x22 46–55. Horizontal and vertical asymptotes xS∞ xS - ∞ a. 1 2 1 2 Evaluate lim f x and lim f x , and then identify any S∞ S - ∞ 1- 16 + 2 -8 x x 21. lim 3x 2 22. lim 2x horizontal asymptotes. xS - ∞ xS - ∞ b. Find the vertical asymptotes. For each vertical asymptote x = a, 1- -52 1 -8 + 32 1 2 1 2 23. lim 12x 24. lim 2x 4x S S evaluate lim- f x and lim+ f x . x ∞ x - ∞ xSa xSa 2 - + 3 + 2 + 1 2 1 2 x 4x 3 2x 10x 12x 1 2 = 1 2 = 25–34. Rational functions Evaluate lim f x and lim f x for the 46. f x 47. f x xS∞ xS - ∞ x - 1 x3 + 2x2 following rational functions. Then give the horizontal asymptote of f 2 4 + 2 + 2 (if any). 1 2 = 16x 64x x 48. f x 2 - 2 - 2x 4 1 2 = 4x 1 2 = 3x 7 25. f x 26. f x + 2 + 4 + 3 - 2 20x 1 x 5x 1 2 = 3x 3x 36x 49. f x 4 2 2 - + 2 - x - 25x + 144 1 2 = 6x 9x 8 1 2 = 4x 7 27. f x 28. f x 2 + 2 + + 1 2 = 2 1 2 - 2 4 + 2 3x 2 8x 5x 2 50. f x 16x 4x 16x 1 3 - 4 + 2 - 1 2 = 3x 7 1 2 = x 7 x 9 1 2 = 29. f x 30. f x 51. f x x4 + 5x2 x5 + x2 - x x1x - 32 + 8 - - 1 2 = 2x 1 1 2 = 12x 3 x 1 1 2 = 31. f x 32. f x 52. f x 3x4 - 2 3x8 - 2x7 x2>3 - 1

5 + 2 - 3 + 2 1 2 = 40x x 1 2 = x 1 2x + 2x + 6 - 3 33. f x 34. f x 1 2 = 53. f x 16x4 - 2x 2x + 8 x - 1

2.5 Limits at Infinity 105

2 1 2 = ∞ 1 2 = - ∞ 1 2 = 1 - x 61. lim f x , lim f x , lim f x 1, 1 2 = xS0+ xS0- xS∞ 54. f x 1 + 2 1 2 = - x x 1 lim f x 2 xS - ∞ 1 2 = 2 - 2 - 55. f x x x 1 62. Asymptotes Find the vertical and horizontal asymptotes of 1 2 = 1>x f x e . 56–59. End behavior for transcendental functions 1 2 = 63. Asymptotes Find the vertical and horizontal asymptotes of 56. The central branch of f x tan x is shown in the figure. + 1 1 2 = cos x 2 x f x . a. Evaluate lim -tan x and lim + tan x. Are these 1 xSp>2 xS-p>2 x infinite limits or limits at infinity? b. Sketch a graph of g1x2 = tan-1x by reflecting the graph of Applications f over the line y = x, and use it to evaluate lim tan-1x and 64–69. Steady states If a function f represents a system that varies in - xS∞ 1 2 1 time, the existence of lim f t means that the system reaches a steady lim tan x. S∞ S - ∞ t x state (or equilibrium). For the following systems, determine if a steady state exists and give the steady-state value. y f (x) tan x 2500 64. The population of a bacteria culture is given by p1t2 = . t + 1 3500t 65. The population of a culture of tumor cells is given by p1t2 = . t + 1 1 66. The amount of drug (in milligrams) in the blood after an IV tube 1 2 = 1 - -t2 q q x is inserted is m t 200 1 2 . 1 67. The value of an investment in dollars is given by v1t2 = 1000e0.065t. 68. The population of a colony of squirrels is given by 1500 p1t2 = . 3 + 2e-0.1t + 1 2 = at sin t b 69. The amplitude of an oscillator is given by a t 2 . T 57. Graph y = sec-1 x and evaluate the following limits using the t graph. Assume the domain is 5x: x Ú 16. a. lim sec-1 x b. lim sec-1 x 70–73. Looking ahead to sequences A sequence is an infinite, ordered xS∞ xS - ∞ list of numbers that is often defined by a function. For example, the 5 c6 1 2 = 58. The hyperbolic cosine function, denoted cosh x, is used to model sequence 2, 4, 6, 8, is specified by the function f n 2n, = c 1 2 the shape of a hanging cable (a telephone wire, for example). It is where n 1, 2, 3, . The limit of such a sequence is lim f n , nS∞ ex + e-x provided the limit exists. All the limit laws for limits at infinity may defined as cosh x = . 2 be applied to limits of sequences. Find the limit of the following sequences, or state that the limit does not exist. a. Determine its end behavior by evaluating lim cosh x and S∞ lim cosh x. x 4 4 2 4 S - ∞ e cf 1 2 = x 70. 4, 2, , 1, , , , which is defined by f n , for b. Evaluate cosh 0. Use symmetry and part (a) to sketch a plau- 3 5 3 n = c sible graph for y = cosh x. n 1, 2, 3, x - -x - e e e 1 2 3 cf 1 2 = n 1 59. The hyperbolic sine function is defined as sinh x = . 71. 0, , , , , which is defined by f n , 2 2 3 4 n = c a. Determine its end behavior by evaluating lim sinh x and for n 1, 2, 3, xS∞ lim sinh x. 1 4 9 16 n2 xS - ∞ e cf 1 2 = 72. , , , , , which is defined by f n , b. Evaluate sinh 0. Use symmetry and part (a) to sketch a plau- 2 3 4 5 n + 1 sible graph for y = sinh x. for n = 1, 2, 3, c 60–61. Sketching graphs Sketch a possible graph of a function f that + e 3 4 5 cf 1 2 = n 1 73. 2, , , , , which is defined by f n , satisfies all the given conditions. Be sure to identify all vertical and 4 9 16 n2 horizontal asymptotes. for n = 1, 2, 3, c 1- 2 = - 1 2 = 1 2 = 1 2 = 60. f 1 2, f 1 2, f 0 0, lim f x 1, xS∞ 1 2 = - lim f x 1 xS - ∞

106 Chapter 2 r Limits

Additional Exercises T 77. Subtle asymptotes Use analytical methods to identify all the p1x2 ln 19 - x 22 1 2 = 1 2 = 74. End behavior of a rational function Suppose f x asymptotes of f x . Then confirm your results by q1x2 2ex - e -x is a rational function, where locating the asymptotes using a graphing calculator. 1 2 = m + m-1 + + 2 + + p x amx am-1x g a2x a1x a0, 1 2 = n + n-1 + + 2 + + ≠ q x bnx bn-1x g b2x b1x b0, am 0, ≠ QUICK CHECK ANSWERS and bn 0. a = 1 2 = m 1. 10>11, 100>101, 1000>1001, 1 2. p1x2 S - ∞ as a. Prove that if m n, then lim f x . S { ∞ x bn x S ∞ and p1x2 S ∞ as x S - ∞ 3. Horizontal 6 1 2 = 10 1 b. Prove that if m n, then lim f x 0. = = S { ∞ asymptote is y 3 ; vertical asymptote is x 3. x - T 1 2 1 2 10x = ∞ 10x = 10x = 75–76. Limits of exponentials Evaluate lim f x and lim f x . 4. lim e , lim e 0, lim e 0,

xS∞ xS-∞ xS∞ xS - ∞ xS∞ Then state the horizontal asymptote(s) of f . Confirm your findings by lim e-10x = ∞ ➤ S plotting f . x - ∞ x + 2x x + -x 1 2 = 2e 3e 1 2 = 3e e 75. f x 76. f x - e2x + e3x ex + e x

2.6 Continuity

The graphs of many functions encountered in this text contain no holes, jumps, or breaks. = 1 2 For example, if L f t represents the length of a fish t years after it is hatched, then the = 1 2 length of the fish changes gradually as t increases. Consequently, the graph of L f t contains no breaks (Figure 2.42a). Some functions, however, do contain abrupt changes in their values. Consider a parking meter that accepts only quarters and each quarter buys 15 minutes of parking. Letting c1t2 be the cost (in dollars) of parking for t minutes, the graph of c has breaks at integer multiples of 15 minutes (Figure 2.42b).

L y

25 1.25

1.00 y ϭ c(t) L ϭ f (t) 0.75

0.50 Length (in) Cost (dollars) 5 0.25

0 1 2 3 4 t 0 15 30 45 60 t Time (yr) Time (min) (a) (b) FIGURE 2.42

QUICK CHECK 1 For what values of t in Informally, we say that a function f is continuous at a if the graph of f contains no

10, 602 does the graph of y = c1t2 in holes or breaks at a (that is, if the graph near a can be drawn without lifting the pencil). If Figure 2.42b have a discontinuity? ➤ a function is not continuous at a, then a is a point of discontinuity.

Continuity at a Point This informal description of continuity is sufficient for determining the continuity of simple functions, but it is not precise enough to deal with more complicated functions such as 1 x sin if x ≠ 0 h1x2 = W x 0 if x = 0.

2.6 Continuity 107

It is difficult to determine whether the graph of h has a break at 0 because it oscillates rap- idly as x approaches 0 (Figure 2.43). We need a better definition.

y Is h continuous 0.2 at x 0?

0.1

0.2 0.2 x 1 0.1 x sin if x 0 h(x) x 0 if x 0 0.2

FIGURE 2.43

DEFINITION Continuity at a Point 1 2 = 1 2 A function f is continuous at a if lim f x f a . If f is not continuous at a, then a S is a point of discontinuity. x a

1 2 = 1 2 1 2 There is more to this definition than first appears. If lim f x f a , then f a and xSa 1 2 lim f x must both exist, and they must be equal. The following checklist is helpful in xSa determining whether a function is continuous at a.

Continuity Checklist In order for f to be continuous at a, the following three conditions must hold. 1 2 1 2 1. f a is defined a is in the domain of f . 1 2 2. lim f x exists. xSa y 1 2 = 1 2 1 2 3. lim f x f a the value of f equals the limit of f at a . xSa 5 y f (x)

3 If any item in the continuity checklist fails to hold, the function fails to be continuous at a. From this definition, we see that continuity has an important practical consequence: 1 2 = 1 2 1 If f is continuous at a, then lim f x f a , and direct substitution may be used to xSa 1 2 evaluate lim f x . xSa 0 1 3 5 7 x

FIGURE 2.44 EXAMPLE 1 Points of discontinuity Use the graph of f in Figure 2.44 to identify values of x on the interval 10, 72 at which f has a discontinuity. ➤ In Example 1, the discontinuities at x = 1 and x = 2 are called removable SOLUTION The function f has discontinuities at x = 1, 2, 3, and 5 because the graph discontinuities because they can be contains holes or breaks at each of these locations. These claims are verified using the removed by redefining the function continuity checklist. 1 2 = at these points (in this case f 1 3 1 2 = 1 2 and f 2 1). The discontinuity at r f 1 is not defined. x = 3 is called a jump discontinuity. 1 2 = 1 2 = 1 2 1 2 r f 2 3 and lim f x 1. Therefore, f 2 and lim f x exist but are not equal. The discontinuity at x = 5 is called an xS2 xS2 infinite discontinuity. These terms are discussed in Exercises 95–101.

108 Chapter 2 r Limits

1 2 1 2 = r lim f x does not exist because the left-sided limit lim f x 2 differs from the xS3 xS3- 1 2 = right-sided limit lim f x 1.

xS3+ 1 2 1 2 ➤ r Neither lim f x nor f 5 exists. Related Exercises 9–12 xS5

EXAMPLE 2 Identifying discontinuities Determine whether the following functions are continuous at a. Justify each answer using the continuity checklist. 2 + + 1 2 = 3x 2x 1 = a. f x ; a 1 x - 1 3x 2 + 2x + 1 b. g1x2 = ; a = 2 x - 1 1 x sin if x ≠ 0 c. h1x2 = W x ; a = 0

0if x = 0 SOLUTION 1 2 a. The function f is not continuous at 1 because f 1 is undefined. b. Because g is a rational function and the denominator is nonzero at 2, it follows by 1 2 = 1 2 = Theorem 2.3 that lim g x g 2 17. Therefore, g is continuous at 2. xS2 c. By definition, h102 = 0. In Exercise 55 of Section 2.3, we used the Squeeze Theorem 1 = 1 2 = 1 2 to show that lim x sin 0. Therefore, lim h x h 0 , which implies that h is

xS0 x xS0 continuous at 0. Related Exercises 13–20 ➤

The following theorems make it easier to test various combinations of functions for continuity at a point.

THEOREM 2.9 Continuity Rules If f and g are continuous at a, then the following functions are also continuous at a. Assume c is a constant and n 7 0 is an integer. a. f + g b. f - g c. cf d. fg > 1 2 ≠ 1 1 22n e. f g, provided g a 0 f. f x

1 2 = 1 2 To prove the first result, note that if f and g are continuous at a, then lim f x f a xSa 1 2 = 1 2 and lim g x g a . From the limit laws of Theorem 2.3, it follows that xSa 1 1 2 + 1 22 = 1 2 + 1 2 lim f x g x f a g a . xSa Therefore, f + g is continuous at a. Similar arguments lead to the continuity of differ- ences, products, quotients, and powers of continuous functions. The next theorem is a direct consequence of Theorem 2.9.

2.6 Continuity 109

THEOREM 2.10 Polynomial and Rational Functions a. A polynomial function is continuous for all x. p b. A rational function (a function of the form , where p and q are polynomials) q is continuous for all x for which q1x2 ≠ 0. x f (x) x2 7x 12 y

20 EXAMPLE 3 Applying the continuity theorems For what values of x is the function 1 2 = x f x continuous? x 2 - 7x + 12 SOLUTION 2 6 x a. Because f is rational, Theorem 2.10b implies it is continuous for all x at which the denominator is nonzero. The denominator factors as 1x - 321x - 42, so it is zero at

20 x = 3 and x = 4. Therefore, f is continuous for all x except x = 3 and x = 4 ( Figure 2.45). Related Exercises 21–26 ➤ Continuous everywhere except x 3 and x 4 The following theorem allows us to determine when a composition of two functions is FIGURE 2.45 continuous at a point. Its proof is informative and is outlined in Exercise 102.

THEOREM 2.11 Continuity of Composite Functions at a Point If g is continuous at a and f is continuous at g1a2, then the composite function f ∘ g is continuous at a.

Theorem 2.11 is useful because it allows us to conclude that the composition of QUICK CHECK 2 Evaluate two continuous functions is continuous at a point. For example, the composite function lim 2x 2 + 9 and 2lim 1x 2 + 92. x 3 S S a b ≠ x 4 x 4 - is continuous for all x 1. The theorem also says that under the stated condi- How do these results illustrate that the x 1 order of a function evaluation and a tions on f and g, the limit of their composition is evaluated by direct substitution; that is,

limit may be switched for continuous 1 1 22 = 1 1 22

➤ lim f g x f g a . functions? xSa

x 4 - 2x + 2 10 EXAMPLE 4 Limit of a composition Evaluate lim a b . xS0 x 6 + 2x 4 + 1 x 4 - 2x + 2 SOLUTION The rational function is continuous for all x because its x 6 + 2x 4 + 1 x 4 - 2x + 2 10 denominator is always positive (Theorem 2.10b). Therefore, a b , which x 6 + 2x 4 + 1 1 2 = 10 is the composition of the continuous function f x x and a continuous rational function, is continuous for all x by Theorem 2.11. By direct substitution, 4 - + 10 4 - # + 10 a x 2x 2 b = a 0 2 0 2 b = 10 = lim # 2 1024.

xS0 x 6 + 2x 4 + 1 06 + 2 04 + 1 Related Exercises 27–30 ➤

Closely related to Theorem 2.11 are two results dealing with limits of composite functions; they are used frequently in upcoming chapters. We present these two results—one a more general version of the other—in a single theorem.

110 Chapter 2 r Limits

THEOREM 2.12 Limits of Composite Functions 1. If g is continuous at a and f is continuous at g1a2, then

1 1 22 = 1 2 lim f g x f 1limg x 2. xSa xSa 2. If limg1x2 = L and f is continuous at L, then xSa 1 1 22 = 1 2 lim f g x f 1limg x 2. xSa xSa

Proof: The first statement follows directly from Theorem 2.11, which states that 1 1 22 = 1 1 22 1 2 = 1 2 lim f g x f g a . If g is continuous at a, then lim g x g a , and it follows that xSa xSa 1 1 22 = 1 1 22 = 1 2 lim f g x f g a f limg x . S 1 S 2

x a b x a 1 2 lim g x xSa

The proof of the second statement relies on the formal definition of a limit, which is discussed in Section 2.7. ➤ Both statements of Theorem 2.12 justify interchanging the order of a limit and a function evaluation. By the second statement, the inner function of the composition needn’t be continuous at the point of interest, but it must have a limit at that point.

EXAMPLE 5 Limits of composite functions Evaluate the following limits. x 2 - 4 a. lim 22x 2 - 1 b. lim cos a b xS-1 xS2 x - 2 SOLUTION a. We show later in this section that 1x is continuous for x Ú 0. The inner function of the composite function 22x 2 - 1 is 2x 2 - 1 and it is continuous and positive at -1. By the first statement of Theorem 2.12,

lim 22x 2 - 1 = lim 2x 2 - 1 = 11 = 1. S S x -1 Ax -1 e 1 b. We show later in this section that cos x is continuous at all points of its domain. The x 2 - 4 x 2 - 4 inner function of the composite function cos a b is , which is not x - 2 x - 2 continuous at 2. However, x 2 - 4 1x - 221x + 22 lim a b = lim = lim1x + 22 = 4. xS2 x - 2 xS2 x - 2 xS2 Therefore, by the second statement of Theorem 2.12, x 2 - 4 x 2 - 4 lim cos a b = cos alima bb = cos 4 ≈ -0.654. S - S - x 2 x 2 x 2 x 2 g

4 Related Exercises 31–34 ➤

2.6 Continuity 111

y Continuity on an Interval Continuous on [a, b) A function is continuous on an interval if it is continuous at every point in that interval. Consider the functions f and g whose graphs are shown in Figure 2.46. Both these functions are continuous for all x in 1a, b2, but what about the endpoints? To answer this ques- y f (x) tion, we introduce the ideas of left-continuity and right-continuity.

DEFINITION Continuity at Endpoints O x 1 2 = 1 2 a b A function f is continuous from the left (or left-continuous) at a if lim f x f a xSa- (a) 1 2 = 1 2 and f is continuous from the right (or right-continuous) at a if lim f x f a . xSa+ y Continuous on (a, b] Combining the definitions of left-continuous and right-continuous with the definition of continuity at a point, we define what it means for a function to be continuous on an y g(x) interval.

DEFINITION Continuity on an Interval A function f is continuous on an interval I if it is continuous at all points of I. If I O a b x contains its endpoints, continuity on I means continuous from the right or left at the (b) endpoints. FIGURE 2.46

To illustrate these definitions, consider again the functions in Figure 2.46. In Figure 2.46a, 1 2 = 1 2 f is continuous from the right at a because lim f x f a ; but it is not continuous from xSa+ 1 2 3 2 the left at b because f b is not defined. Therefore, f is continuous on the interval a, b . The behavior of the function g in Figure 2.46b is the opposite: It is continuous from the left at b, but it is not continuous from the right at a. Therefore, g is continuous on 1a, b4.

QUICK CHECK 3 Modify the graphs of the functions f and g in Figure 2.46 to obtain functions that are continuous on 3a, b4. ➤

EXAMPLE 6 Intervals of continuity Determine the intervals of continuity for 2 + … 1 2 = e x 1 if x 0 f x y 3x + 5 if x 7 0.

10 SOLUTION This piecewise function consists of two polynomials that describe a parabola and a line (Figure 2.47). By Theorem 2.10, f is continuous for all x ≠ 0. From its graph, y f (x) Continuous it appears that f is left-continuous at 0. This observation is verified by noting that on (0, )

1 2 = 1 2 + 2 = lim f x lim x 1 1, xS0- xS0- 1 2 = 1 2 which means that lim- f x f 0 . However, because Continuous 2 Left-continuous xS0 on (, 0] at x 0 1 2 = 1 + 2 = ≠ 1 2 lim f x lim 3x 5 5 f 0 , S + S + 2 2 x x 0 x 0 we see that f is not right-continuous at 0. Therefore, we can also say that f is continuous

FIGURE 2.47 on 1- ∞, 04 and on 10, ∞2. Related Exercises 35–40 ➤

112 Chapter 2 r Limits

Functions Involving Roots Recall that Limit Law 7 of Theorem 2.3 states 3 1 24 n>m = 1 2 n>m lim f x 3lim f x 4 , xSa xSa 1 2 Ú > provided f x 0, for x near a, if m is even and n m is reduced. Therefore, if m is odd 3 1 24 n>m and f is continuous at a, then f x is continuous at a, because 3 1 24 n>m = 1 2 n>m = 3 1 24 n>m lim f x 3lim f x 4 f a . xSa xSa 3 1 24 n>m When m is even, the continuity of f x must be handled more carefully because 1 2 Ú this function is defined only when f x 0. Exercise 59 of Section 2.7 establishes an important fact: If f is continuous at a and f 1a2 7 0, then f is positive for all values of x in the domain sufficiently close to a. Combining this fact with Theorem 2.11 (the continuity of composite functions), it fol- 3 1 24 n>m 1 2 7 1 2 = lows that f x is continuous at a provided f a 0. At points where f a 0, the 3 1 24 n>m 3 1 24 n>m behavior of f x varies. Often we find that f x is left- or right-continuous at that point, or it may be continuous from both sides.

THEOREM 2.13 Continuity of Functions with Roots Assume that m and n are positive integers with no common factors. If m is an odd 3 1 24 n>m integer, then f x is continuous at all points at which f is continuous. 3 1 24 n>m If m is even, then f x is continuous at all points a at which f is continuous 1 2 7 and f a 0.

Continuous on [3, 3] EXAMPLE 6 Continuity with roots For what values of x are the following functions y continuous? 1 2 = 2 - 2 1 2 = 1 2 - + 22>3 a. g x 9 x b. f x x 2x 4 4 g(x) ͙9 x2 SOLUTION a. The graph of g is the upper half of the circle x 2 + y2 = 9 (which can be verified 2 by solving x 2 + y2 = 9 for y). From Figure 2.48, it appears that g is continuous on 3-3, 34. To verify this fact, note that g involves an even root 1m = 2, n = 1 in Theorem 2.13). If -3 6 x 6 3, then 9 - x 2 7 0 and by Theorem 2.13, g is 1- 2 3 3 x continuous for all x on 3, 3 . At the right endpoint, lim 29 - x 2 = 0 = g132 by Limit Law 7, which implies Right-continuous Left-continuous xS3- at x 3 at x 3 that g is left-continuous at 3. Similarly, g is right-continuous at -3 because FIGURE 2.48 lim 29 - x 2 = 0 = g1-32. Therefore, g is continuous on 3-3, 34. xS - 3 + b. The polynomial x 2 - 2x + 4 is continuous for all x by Theorem 2.10a. Because f

QUICK CHECK 4 On what interval is involves an odd root (m = 3, n = 2 in Theorem 2.13), f is continuous for all x. 1 2 = 1>4 Related Exercises 41–50 ➤

f x x continuous? On what 1 2 = 2>5 ➤ interval is f x x continuous? Continuity of Transcendental Functions The understanding of continuity that we have developed with algebraic functions may now be applied to transcendental functions.

Trigonometric Functions In Example 8 of Section 2.3, we used the Squeeze Theorem = = = = to show that lim sin x 0 and lim cos x 1. Because sin 0 0 and cos 0 1, these xS0 xS0

2.6 Continuity 113

limits imply that sin x and cos x are continuous at 0. The graph of y = sin x (Figure 2.49) = suggests that lim sin x sin a for any value of a, which means that sin x is continuous xSa everywhere. The graph of y = cos x also indicates that cos x is continuous for all x. Exercise 105 outlines a proof of these results. With these facts in hand, we appeal to Theorem 2.9e to discover that the remain- ing trigonometric functions are continuous on their domains. For example, because sec x = 1>cos x, the secant function is continuous for all x for which cos x ≠ 0 (for all x except odd multiples of p>2) (Figure 2.50). Likewise, the tangent, cotangent, and cose- cant functions are continuous at all points of their domains.

ϭ y y sec x

y 3 y ϭ sin x 1

(a, sin a) x sin a ... sin x ៬ sin a Ϫw Ϫq qw

Ϫ3

x a sec x is continuous at all As x ៬ a... points of its domain. FIGURE 2.49 FIGURE 2.50

y Exponential Functions The continuity of exponential functions of the form 1 2 = x 6 6 7 x f x b , with 0 b 1 or b 1, raises an important question. Consider the function f (x) 4 1 2 = x f x 4 (Figure 2.51). Evaluating f is routine if x is rational:

3 = # # = -2 = 1 = 1 3>2 = 2 3 = -1>3 = 1 4 4 4 4 64; 4 2 ; 4 4 8; and 4 . 4 16 23 4 But what is meant by 4x when x is an irrational number, such as 12? In order for 1 2 = x (0, 1) f x 4 to be continuous for all real numbers, it must also be defined when x is an irrational number. Providing a working definition for an expression such as 412 requires x mathematical results that don’t appear until Chapter 6. Until then, we assume without 1 2 = x Exponential functions are defined proof that the domain of f x b is the set of all real numbers and that f is continuous for all real numbers and are at all points of its domain. continuous on (, ), as shown in Chapter 6 Inverse Functions Suppose a function f is continuous and one-to-one on an inter- FIGURE 2.51 val I. Reflecting the graph of f through the line y = x generates the graph of f -1. The reflection process introduces no discontinuities in the graph of f -1, so it is plausible (and indeed, true) that f -1 is continuous on the interval corresponding to I. We state this fact without a formal proof.

THEOREM 2.14 Continuity of Inverse Functions If a continuous function f has an inverse on an interval I, then its inverse f -1 is 1 2 also continuous (on the interval consisting of the points f x , where x is in I).

Because all the trigonometric functions are continuous on their domains, they are also continuous when their domains are restricted for the purpose of defining inverse functions. Therefore, by Theorem 2.14, the inverse trigonometric functions are continuous at all points of their domains.

114 Chapter 2 r Limits

1 2 = Logarithmic functions of the form f x logb x are continuous at all points of their domains for the same reason: They are inverses of exponential functions, which are one-to-one and continuous. Collecting all these facts together, we have the following theorem.

THEOREM 2.15 Continuity of Transcendental Functions The following functions are continuous at all points of their domains. Trigonometric Inverse Trigonometric Exponential sin x cos x sin-1 x cos-1 x bx ex tan x cot x tan-1 x cot-1 x Logarithmic -1 -1 sec x csc x sec x csc x logb x ln x

1 2 = 1 2 For each function listed in Theorem 2.15, we have lim f x f a , provided a is in xSa the domain of the function. This means that limits involving these functions may be evaluated by direct substitution at points in the domain.

EXAMPLE 7 Limits involving transcendental functions Evaluate the following limits after determining the continuity of the functions involved. cos2 x - 1 a. lim b. lim 1 24 ln x + tan-1 x2 xS0 cos x - 1 xS1 SOLUTION

➤ Limits like the one in Example 7a a. Both cos2 x - 1 and cos x - 1 are continuous for all x by Theorems 2.9 and 2.15. are denoted 0/0 and are known as However, the ratio of these functions is continuous only when cos x - 1 ≠ 0, which indeterminate forms, to be studied occurs when x is not an integer multiple of 2p. Note that both the numerator and denominator further in Section 4.7. cos2 x - 1 of approach 0 as x S 0. To evaluate the limit, we factor and simplify: cos x - 1 cos2 x - 1 1cos x - 121cos x + 12 lim = lim = lim 1cos x + 12 xS0 cos x - 1 xS0 cos x - 1 xS0 (where cos x - 1 may be canceled because it is nonzero as x approaches 0). The limit on the right is now evaluated using direct substitution: 1 + 2 = + = lim cos x 1 cos 0 1 2. xS0 b. By Theorem 2.15, ln x is continuous on its domain 10, ∞2. However, ln x 7 0 only when x 7 1, so Theorem 2.13 implies 24 ln x is continuous on 11, ∞2. At x = 1, 24 ln x is right-continuous (Quick Check 5). The domain of tan-1 x is all real 1- ∞ ∞2 1 2 = 24 + -1 numbers, and it is continuous on , . Therefore, f x ln x tan x is 3 ∞2 6 QUICK CHECK 5 Show that continuous on 1, . Because the domain of f does not include points with x 1, 4 - 4 - 4 1 2 + 1 2 1 2 + 1 2

1 2 = 2 lim- ln x tan x does not exist, which implies that lim ln x tan x f x ln x is right-continuous xS1 xS1

➤ at x = 1. does not exist. Related Exercises 51–56 ➤

We close this section with an important theorem that has both practical and theoretical uses.

The Intermediate Value Theorem 1 2 = A common problem in mathematics is finding solutions to equations of the form f x L. Before attempting to find values of x satisfying this equation, it is worthwhile to determine whether a solution exists.

2.6 Continuity 115

The existence of solutions is often established using a result known as the Intermediate 1 2 Value Theorem. Given a function f and a constant L, we assume L lies between f a and 1 2 3 4 f b . The Intermediate Value Theorem says that if f is continuous on a, b , then the graph of f must cross the horizontal line y = L at least once (Figure 2.52). Although this theorem is easily illustrated, its proof goes beyond the scope of this text.

Intermediate Value Theorem y y

f (b) y f (x) y f (x) f (a) L L

f (b) f (a)

O x O x a bc a c1 c2 c3 b In (a, b), there is at least one number c such that f (c) L, where L is between f (a) and f (b).

FIGURE 2.52 y f is not continuous on [a, b]... THEOREM 2.16 The Intermediate Value Theorem 3 4 f (b) Suppose f is continuous on the interval a, b and L is a number strictly between y ϭ f (x) 1 2 1 2 1 2 1 2 = f a and f b . Then there exists at least one number c in a, b satisfying f c L.

L The importance of continuity in Theorem 2.16 is illustrated in Figure 2.53, where we see a function f that is not continuous on 3a, b4. For the value of L shown in the figure, f (a) 1 2 1 2 = there is no value of c in a, b satisfying f c L. The next example illustrates a practical application of the Intermediate Value Theorem. O a b x EXAMPLE 8 Finding an interest rate Suppose you invest $1000 in a special 5-year ... and there is no number c in (a, b) such that f (c) ϭ L. savings account with a fixed annual interest rate r, with monthly compounding. The r 60 amount of money A in the account after 5 years (60 months) is A1r2 = 1000a1 + b . FIGURE 2.53 12 Your goal is to have $1400 in the account after 5 years. QUICK CHECK 6 Does the equation a. Use the Intermediate Value Theorem to show there is a value of r in (0, 0.08)—that is,

1 2 = 3 + + = f x x x 1 0 have a solu- 1 2 = ➤ an interest rate between 0% and 8%—for which A r 1400. tion on the interval 3-1, 14? Explain. b. Use a graphing utility to illustrate your explanation in part (a), and then estimate the y interest rate required to reach your goal. SOLUTION 2000 y ϭ A(r) r 60 ϭ a. As a polynomial in r (of degree 60), A1r2 = 1000a1 + b is continuous y 1400 12 for all r. Evaluating A1r2 at the endpoints of the interval 30, 0.084, we have Interest rate that yields A102 = 1000 and A10.082 ≈ 1489.85. Therefore,

Amount of money 500 $1400 after 5 years (dollars) after 5 years A102 6 1400 6 A10.082, 0 0.02 0.0675 0.10 r Interest rate and it follows, by the Intermediate Value Theorem, that there is a value of r 1 2 1 2 = FIGURE 2.54 in 0, 0.08 for which A r 1400. b. The graphs of y = A1r2 and the horizontal line y = 1400 are shown in Figure 2.54; it is evident that they intersect between r = 0 and r = 0.08. Solving A1r2 = 1400 algebraically or using a root finder reveals that the curve and line intersect at r ≈ 0.0675. Therefore, an interest rate of approximately 6.75% is required for the investment to be

worth $1400 after 5 years. Related Exercises 57–64 ➤

116 Chapter 2 r Limits

SECTION 2.6 EXERCISES Review Questions 13–20. Continuity at a point Determine whether the following 1. Which of the following functions are continuous for all values in functions are continuous at a. Use the continuity checklist to their domain? Justify your answers. justify your answer. a. a1t2 = altitude of a skydiver t seconds after jumping from a 2 + + 1 2 = 2x 3x 1 = plane 13. f x ; a 5 x 2 + 5x b. n1t2 = number of quarters needed to park in a metered park- 2 + + ing space for t minutes 1 2 = 2x 3x 1 = - 14. f x ; a 5 c. T1t2 = temperature t minutes after midnight in Chicago on x 2 + 5x January 1 1 2 = 1 - = d. p1t2 = number of points scored by a basketball player after 15. f x x 2; a 1 t minutes of a basketball game 1 16. g1x2 = ; a = 3 2. Give the three conditions that must be satisfied by a function to be x - 3 continuous at a point. 2 - x 1 ≠ 1 2 = c - if x 1 = 3. What does it mean for a function to be continuous on an interval? 17. f x x 1 ; a 1 = 4. We informally describe a function f to be continuous at a if its 3if x 1 graph contains no holes or breaks at a. Explain why this is not an 2 - + x 4x 3 ≠ adequate definition of continuity. 1 2 = c - if x 3 = 18. f x x 3 ; a 3

5. Complete the following sentences. 2if x = 3 a. A function is continuous from the left at a if ______. - 1 2 = 5x 2 = 19. f x ; a 4 b. A function is continuous from the right at a if ______. x 2 - 9x + 20 6. Describe the points (if any) at which a rational function fails to be 2 + x x ≠ - continuous. 1 2 = c + if x 1 = - 20. f x x 1 ; a 1 1 2 = x> 7. What is the domain of f x e x and where is f continuous? 2if x = -1 8. Explain in words and pictures what the Intermediate Value Theo- 21–26. Continuity on intervals Use Theorem 2.10 to determine the rem says. intervals on which the following functions are continuous. 3x 2 - 6x + 7 Basic Skills 21. p1x2 = 4x 5 - 3x 2 + 1 22. g1x2 = 9–12. Discontinuities from a graph Determine the points at which x 2 + x + 1 the following functions f have discontinuities. For each point, state the 5 + + 2 - + 1 2 = x 6x 17 1 2 = x 4x 3 conditions in the continuity checklist that are violated. 23. f x 24. s x x 2 - 9 x 2 - 1 y 10. y 9. + 1 2 = 1 1 2 = t 2 5 25. f x 26. f t 5 x 2 - 4 t2 - 4 4 4 y ϭ f (x) y ϭ f (x) 27–30. Limits of compositions Evaluate the following limits and 3 3 justify your answer. 4 2 3 2 27. lim 1x 8 - 3x 6 - 1240 28. lim a b xS0 xS2 5 - 2 - 1 1 2x 4x 50 x + 5 4 2x + 1 3 0 x a b a b 0 12345 x 12345 29. lim 30. lim xS1 x + 2 xS∞ x 11. y 12. y 31–34. Limits of composite functions Evaluate the following limits and justify your answer. 5 5 ϭ ϭ y f (x) y f (x) x 3 - 2x 2 - 8x t - 4 4 4 31. lim 32. lim tan S B - S x 4 x 4 t 4 2t - 2 3 3 1>3 a sin x b a x b 2 2 33. lim ln 2 34. lim xS0 x xS0 216x + 1 - 1 1 1 35–38. Intervals of continuity Determine the intervals of continuity 0 x 0 12345 x for the following functions. 12345 35. The graph of Exercise 9 36. The graph of Exercise 10

2.6 Continuity 117

37. The graph of Exercise 11 38. The graph of Exercise 12 b. Use a graph to illustrate your explanation in part (a); then approximate the interest rate required to reach your goal. 39. Intervals of continuity Let T 58. Intermediate Value Theorem and mortgage payments You 2 + Ú are shopping for a $150,000, 30-year (360-month) loan to buy a 1 2 = e x 3x if x 1 f x 2x if x 6 1. house. The monthly payment is 150,0001r>122 a. Use the continuity checklist to show that f is not continuous at 1. m1r2 = , 1 - 11 + r>122-360 b. Is f continuous from the left or right at 1? c. State the interval(s) of continuity. where r is the annual interest rate. Suppose banks are currently offering interest rates between 6% and 8%. 40. Intervals of continuity Let a. Use the Intermediate Value Theorem to show there is a value 3 + + … 1 2 = e x 4x 1if x 0 of r in (0.06, 0.08)—an interest rate between 6% and 8%—that f x 3 2x if x 7 0. allows you to make monthly payments of $1000 per month. b. Use a graph to illustrate your explanation to part (a). Then a. Use the continuity checklist to show that f is not continuous at 0. determine the interest rate you need for monthly payments of b. Is f continuous from the left or right at 0? $1000. c. State the interval(s) of continuity. T 59–64. Applying the Intermediate Value Theorem 41–46. Functions with roots Determine the interval(s) on which the following functions are continuous. Be sure to consider right- and a. Use the Intermediate Value Theorem to show that the following left-continuity at the endpoints. equations have a solution on the given interval. b. 1 2 = 2 2 - 1 2 = 2 4 - Use a graphing utility to find all the solutions to the equation on 41. f x 2x 16 42. g x x 1 the given interval. 1 2 = 23 2 - - 1 2 = 1 2 - 23>2 43. f x x 2x 3 44. f t t 1 c. Illustrate your answers with an appropriate graph. 1 2 = 1 - 22>3 1 2 = 1 - 23>4 45. f x 2x 3 46. f z z 1 59. 2x 3 + x - 2 = 0; 1-1, 12 47–50. Limits with roots Determine the following limits and justify 60. 2x 4 + 25x 3 + 10 = 5; 10, 12 your answers. 3 - 2 + = - 1- 2 + 61. x 5x 2x 1; 1, 5 4x 10 1 2 - + 23 2 - 2 47. lim 48. lim x 4 x 9 - 5 - 2 + 1 + = 1 2 xS2 A 2x - 2 xS -1 62. x 4x 2 x 5 0; 0, 3 x t2 + 5 63. x + e = 0; 1-1, 02 49. lim 1 2x 2 + 72 50. lim xS3 tS2 1 + 2t2 + 5 64. x ln x - 1 = 0; 11, e2

51–56. Continuity and limits with transcendental functions Further Explorations Determine the interval(s) on which the following functions are 65. Explain why or why not Determine whether the following state- continuous; then evaluate the given limits. ments are true and give an explanation or counterexample. 1 2 = 1 2 1 2 51. f x csc x; lim f x ; lim f x a. If a function is left-continuous and right-continuous at a, then xSp>4 xS2p- it is continuous at a. 1 2 = 1x 1 2 1 2 52. f x e ; lim f x ; lim f x b. If a function is continuous at a, then it is left-continuous and xS4 xS0+ right-continuous at a. + 6 1 2 … … 1 2 1 2 = 1 sin x 1 2 1 2 c. If a b and f a L f b , then there is some value of c 53. f x ; lim f x ; lim f x S - S 1 2 1 2 = cos x x p>2 x 4p>3 in a, b for which f c L. d. Suppose f is continuous on 3a, b4. Then there is a point c in ln x 1 2 1 2 = 1 1 2 + 1 22> 1 2 = 1 2 a, b such that f c f a f b 2. 54. f x - ; lim f x sin 1 x xS1- 66. Continuity of the absolute value function Prove that the x 1 2 = e 1 2 1 2 absolute value function x is continuous for all values of x. 55. f x x; lim- f x ; lim+ f x 1 - e xS0 xS0 (Hint: Using the definition of the absolute value function, compute lim x and lim x .) 2x - S - S + 1 2 = e 1 1 2 x 0 x 0 56. f x x ; lim f x e - 1 xS0 67–70. Continuity of functions with absolute values Use the continuity of the absolute value function (Exercise 66) to determine T 57. Intermediate Value Theorem and interest rates Suppose $5000 the interval(s) on which the following functions are continuous. is invested in a savings account for 10 years (120 months), with + an annual interest rate of r, compounded monthly. The amount of 1 2 = 2 + - 1 2 = ` x 4 ` 67. f x x 3x 18 68. g x money in the account after 10 years is A1r2 = 500011 + r>122120. x 2 - 4 1 a. Use the Intermediate Value Theorem to show there is a value 69. h1x2 = ` ` 70. h1x2 = x2 + 2x + 5 + 1x of r in (0, 0.08)—an interest rate between 0% and 8%—that 1x - 4 allows you to reach your savings goal of $7000 in 10 years.

118 Chapter 2 r Limits

71–80. Miscellaneous limits Evaluate the following limits. a. Determine the value of a for which g is continuous from the left at 1. cos2 x + 3 cos x + 2 sin2 x + 6 sin x + 5 71. lim 72. lim b. Determine the value of a for which g is continuous from the Sp + S p> 2 - x cos x 1 x 3 2 sin x 1 right at 1. c. Is there a value of a for which g is continuous at 1? Explain. 1 - 1 sin x - 1 2 + sin u 2 T 73. lim 74. lim 86. Asymptotes of a function containing exponentials Let xSp>2 1sin x - 1 uS0 sin u x + 3x 1 2 = 2e 5e 1 2 1 2 1 2 f x . Evaluate lim f x , lim f x , lim f x , 2x - 3x S - S + S-∞ cos x - 1 1 - cos2 x e e x 0 x 0 x 75. lim 76. lim 1 2 S 2 S + and lim f x . Then give the horizontal and vertical asymptotes x 0 sin x x 0 sin x xS∞ of f. Plot f to verify your results. tan-1 x cos t 77. lim 78. lim 3t T 87. Asymptotes of a function containing exponentials Let xS∞ x tS ∞ e x + -x 1 2 = 2e 10e 1 2 1 2 f x - . Evaluate lim f x , lim f x , and x x x + x S S-∞ 79. lim 80. lim e e x 0 x xS1- xS0+ 1 2 ln x ln x lim f x . Then give the horizontal and vertical asymptotes of f. xS∞ 81. Pitfalls using technology The graph of the sawtooth function Plot f to verify your results. y = x - :x;, where :x; is the greatest integer function or floor T 88–89. Applying the Intermediate Value Theorem Use the function (Exercise 37, Section 2.2), was obtained using a graph- Intermediate Value Theorem to verify that the following equations have ing utility (see figure). Identify any inaccuracies appearing in the three solutions on the given interval. Use a graphing utility to find the graph and then plot an accurate graph by hand. approximate roots. ϭ Ϫ : ; y x x 88. x 3 + 10x 2 - 100x + 50 = 0; 1-20, 102 1.5 89. 70x 3 - 87x 2 + 32x - 3 = 0; 10, 12

Applications 90. Parking costs Determine the intervals of continuity for the parking cost function c introduced at the outset of this section Ϫ22 (see figure). Consider 0 … t … 60.

Ϫ0.5 y sin x 1.25 T 1 2 = 82. Pitfalls using technology Graph the function f x using x ϭ a graphing window of 3-p, p4 * 30, 24. 1.00 y c(t) a. Sketch a copy of the graph obtained with your graphing device 0.75 and describe any inaccuracies appearing in the graph. 0.50 Cost (dollars) b. Sketch an accurate graph of the function. Is f continuous at 0? 0.25 sin x c. What is the value of lim . S x 0 x 0 15 30 45 60 t Time (min) 83. Sketching functions a. Sketch the graph of a function that is not continuous at 1, but is 91. Investment problem Assume you invest $250 at the end of each defined at 1. year for 10 years at an annual interest rate of r. The amount of money b. Sketch the graph of a function that is not continuous at 1, but 250111 + r210 - 12 in your account after 10 years is A = . has a limit at 1. r Assume your goal is to have $3500 in your account after 10 years. 84. An unknown constant Determine the value of the constant a for which the function a. Use the Intermediate Value Theorem to show that there is an interest rate r in the interval 10.01, 0.102—between 1% and x 2 + 3x + 2 if x ≠ -1 10%—that allows you to reach your financial goal. 1 2 = c + f x x 1 b. Use a calculator to estimate the interest rate required to reach a if x = -1 your financial goal. 92. Applying the Intermediate Value Theorem Suppose you park is continuous at -1. your car at a trailhead in a national park and begin a 2-hr hike to a 85. An unknown constant Let lake at 7 a.m. on a Friday morning. On Sunday morning, you leave 2 + 6 the lake at 7 a.m. and start the 2-hr hike back to your car. Assume x x if x 1 1 2 the lake is 3 mi from your car. Let f t be your distance from the 1 2 = = g x W a if x 1 car t hours after 7 a.m. on Friday morning and let g1t2 be your dis- 3x + 5if x 7 1. tance from the car t hours after 7 a.m. on Sunday morning.

2.6 Continuity 119

1 2 1 2 1 2 1 2 2 a. Evaluate f 0 , f 2 , g 0 , and g 2 . x - 1 1 2 = 1 2 - 1 2 1 2 1 2 if x ≠ 1 b. Let h t f t g t . Find h 0 and h 2 . 98. g1x2 = c 1 - x c. Use the Intermediate Value Theorem to show that there is ; x = 1 3if x = 1 some point along the trail that you will pass at exactly the same time of morning on both days. 99. Do removable discontinuities exist? Refer to Exercises 95–96. 1 2 = 1 > 2 93. The monk and the mountain A monk set out from a monastery in a. Does the function f x x sin 1 x have a removable = the valley at dawn. He walked all day up a winding path, stopping discontinuity at x 0? 1 2 = 1 > 2 for lunch and taking a nap along the way. At dusk, he arrived at a b. Does the function g x sin 1 x have a removable = temple on the mountaintop. The next day, the monk made the return discontinuity at x 0? walk to the valley, leaving the temple at dawn, walking the same T 100–101. Classifying discontinuities Classify the discontinuities in path for the entire day, and arriving at the monastery in the evening. the following functions at the given points. See Exercises 95–96. Must there be one point along the path that the monk occupied at x - 2 the same time of day on both the ascent and descent? (Hint: The 1 2 = = 100. f x ; x 2 question can be answered without the Intermediate Value Theorem.) x - 2 (Source: Arthur Koestler, The Act of Creation.) x 3 - 4x 2 + 4x 101. h1x2 = ; x = 0 and x = 1 Additional Exercises x1x - 12 f f 94. Does continuity of imply continuity of ? Let 102. Continuity of composite functions Prove Theorem 2.11: If g is 1if x Ú 0 continuous at a and f is continuous at g1a2, then the composition g1x2 = e ∘ -1if x 6 0. f g is continuous at a. (Hint: Write the definition of continuity for f and g separately; then combine them to form the definition a. Write a formula for g1x2 . of continuity for f ∘ g.) b. Is g continuous at x = 0? Explain. c. Is g continuous at x = 0? Explain. 103. Continuity of compositions d. For any function f, if f is continuous at a, does it necessar- a. Find functions f and g such that each function is continuous at ily follow that f is continuous at a? Explain. 0, but f ∘ g is not continuous at 0. b. Explain why examples satisfying part (a) do not contradict 95–96. Classifying discontinuities The discontinuities in graphs Theorem 2.11. (a) and (b) are removable discontinuities because they disappear if we 1 2 = 1 2 define or redefine f at a so that f a lim f x . The function in 104. Violation of the Intermediate Value Theorem? Let S x a x graph (c) has a jump discontinuity because left and right limits exist at 1 2 = 1- 2 = - 1 2 = f x . Then f 2 1 and f 2 1. Therefore, a but are unequal. The discontinuity in graph (d) is an infinite discon- x 1- 2 6 6 1 2 - tinuity because the function has a vertical asymptote at a. f 2 0 f 2 , but there is no value of c between 2 and 2 1 2 = for which f c 0. Does this fact violate the Intermediate Value y y Theorem? Explain. 105. Continuity of sin x and cos x y ϭ f (x) y ϭ f (x) a. Use the identity sin 1a + h2 = sin a cos h + cos a sin h with Removable Removable the fact that lim sin x = 0 to prove that lim sin x = sin a, discontinuity discontinuity xS0 xSa thereby establishing that sin x is continuous for all x. (Hint: Let h = x - a so that x = a + h and note that h S 0 as x S a.) O x O x a a b. Use the identity cos 1a + h2 = cos a cos h - sin a sin h with (a) (b) the fact that lim cos x = 1 to prove that lim cos x = cos a. xS0 xSa y Jump y discontinuity y ϭ f (x) QUICK CHECK ANSWERS y ϭ f (x) = Infinite 1. t 15, 30, 45 2. Both expressions have a value of 5, 1 1 22 = 1 1 22 discontinuity showing that lim f g x f lim g x . 3. Fill xSa xSa in the endpoints. 4. 30, ∞2; 1- ∞, ∞2 5. Note that O a x O a x 24 = 24 = 1 2 = 24 = lim+ ln x lim ln x 0 and f 1 ln 1 0. (c) (d) xS1 xS1+ Because the limit from the right and the value of the 95. Is the discontinuity at a in graph (c) removable? Explain. function at x = 1 are equal, the function is right-continuous 96. Is the discontinuity at a in graph (d) removable? Explain. at x = 1. 6. The equation has a solution on the interval

3-1, 14 because f is continuous on 3-1, 14 and 97–98. Removable discontinuities Show that the following functions 1- 2 6 6 1 2 ➤ have a removable discontinuity at the given point. See Exercises 95–96. f 1 0 f 1 . 2 - + 1 2 = x 7x 10 = 97. f x ; x 2 x - 2

120 Chapter 2 r Limits 2.7 Precise Definitions of Limits

The limit definitions already encountered in this chapter are adequate for most elementary limits. However, some of the terminology used, such as sufficiently close and arbitrarily large, needs clarification. The goal of this section is to give limits a solid mathematical foundation by transforming the previous limit definitions into precise mathematical statements.

Moving Toward a Precise Definition Assume the function f is defined for all x near a, except possibly at a. Recall that 1 2 = 1 2 lim f x L means that f x is arbitrarily close to L for all x sufficiently close (but not xSa equal) to a. This limit definition is made precise by observing that the distance between 1 2 1 2 - - f x and L is f x L and that the distance between x and a is x a . Therefore, we 1 2 = 1 2 - write lim f x L if we can make f x L arbitrarily small for any x, distinct from a, xSa - 1 2 - with x a sufficiently small. For instance, if we want f x L to be less than 0.1, then we must find a number d 7 0 such that ➤ The phrase for all x near a means for all 1 2 - 6 - 6 d ≠ f x L 0.1 whenever x a and x a. x in an open interval containing a. 1 2 - If, instead, we want f x L to be less than 0.001, then we must find another number ➤ The Greek letters d (delta) and e d 7 0 such that (epsilon) represent small positive 1 2 - 6 6 - 6 d numbers when discussing limits. f x L 0.001 whenever 0 x a . e 7 d 7 ➤ The two conditions x - a 6 d For the limit to exist, it must be true that for any 0, we can always find a 0 and x ≠ a are written concisely as such that 0 6 x - a 6 d. 1 2 - 6 e 6 - 6 d f x L whenever 0 x a . y D 7 EXAMPLE 1 Determining values of from a graph Figure 2.55 shows the graph of 1 2 = e 7 a linear function f with lim f x 5. For each value of 0, determine a value of xS3 y f (x) d 7 0 satisfying the statement 5 1 2 - 6 e 6 - 6 d lim f (x) 5 f x 5 whenever 0 x 3 . xᠬ3 3 a. e = 1 e = 1 b. 2 1 SOLUTION

0 x e = 1 2 1 2 1 3 5 7 a. With 1, we want f x to be less than 1 unit from 5, which means f x is between 4 and 6. To determine a corresponding value of d, draw the horizontal lines y = 4 and FIGURE 2.55 y = 6 (Figure 2.56a). Then sketch vertical lines passing through the points where the ➤ The founders of calculus, Isaac Newton horizontal lines and the graph of f intersect (Figure 2.56b). We see that the vertical (1642–1727) and Gottfried Leibniz = = 1 2 lines intersect the x-axis at x 1 and x 5. Note that f x is less than 1 unit from (1646–1716), developed the core ideas 5 on the y-axis if x is within 2 units of 3 on the x-axis. So for e = 1, we let d = 2 or of calculus without using a precise any smaller positive value. definition of a limit. It was not until the 19th century that a rigorous definition was introduced by Louis Cauchy (1789–1857) and later refined by Karl Weierstrass (1815–1897).

2.7 Precise Definitions of Limits 121

y y Values of x such that f (x) 5 1

6 6

f (x) 5 1 5 5 f (x) 5 1

4 4

0 1 3 5 7 x 0 1 3 5 7 x

0 x 3 2 (a) (b) FIGURE 2.56 ➤ d e = 1 1 2 1 2 Once an acceptable value of is found b. With 2, we want f x to lie within a half-unit of 5 or, equivalently, f x must lie satisfying the statement 1 2 between 4.5 and 5.5. Proceeding as in part (a), we see that f x is within a half-unit 1 2 - 6 e of 5 on the y-axis (Figure 2.57a) if x is less than 1 unit from 3 (Figure 2.57b). So for f x L whenever e = 1, we let d = 1 or any smaller positive number. 0 6 x - a 6 d, 2

any smaller positive value of d also works. Values of x such that y y f (x) 5 q

5.5 5.5 q f (x) 5 q 5 5 f (x) 5 4.5 4.5

0 x 0 32 4 x 432 (a) 0 x 3 1

FIGURE 2.57 (b) Values of x such that Related Exercises 9–12 ➤ y f (x) 5 Ω The idea of a limit, as illustrated in Example 1, may be described in terms of a contest 6 between two people named Epp and Del. First, Epp picks a particular number e 7 0; then he challenges Del to find a corresponding value of d 7 0 such that 5Ω 1 2 - 6 e 6 - 6 d 5 f x 5 whenever 0 x 3 . (1) Ω 5 e = f (x) 5 Ω To illustrate, suppose Epp chooses 1. From Example 1, we know that Del will sat- 6 d … e = 1 isfy (1) by choosing 0 2. If Epp chooses 2, then (by Example 1) Del re- 4 6 d … e = 1 6 d … 1 sponds by letting 0 1. If Epp lets 8, then Del chooses 0 4 (Figure 2.58). In fact, there is a pattern: For any e 7 0 that Epp chooses, no mat- 0 2 3 4 x 3~ 3~ ter how small, Del will satisfy (1) by choosing a positive value of d satisfying 0 6 d … 2e. Del has discovered a mathematical relationship: If 0 6 d … 2e and 0 x 3 ~ 6 - 6 d 1 2 - 6 e e 7 0 x 3 , then f x 5 , for any 0. This conversation illustrates the 1 2 = FIGURE 2.58 general procedure for proving that lim f x L. xSa

122 Chapter 2 r Limits

QUICK CHECK 1 In Example 1, find a positive number d satisfying the statement 1 2 - 6 1 6 - 6 d ➤ f x 5 whenever 0 x 3 . 100 A Precise Definition Example 1 dealt with a linear function, but it points the way to a precise definition of a 1 2 = limit for any function. As shown in Figure 2.59, lim f x L means that for any positive xSa number e, there is another positive number d such that 1 2 - 6 e 6 - 6 d f x L whenever 0 x a . In all limit proofs, the goal is to find a relationship between e and d that gives an admissi- ble value of d, in terms of e only. This relationship must work for any positive value of e.

lim f (x) L xᠬa y

L ⑀

f (x) L ... then f (x) L ⑀.

L ⑀ y f (x)

O ␦ x aa a ␦ x

If 0 x a ␦...

FIGURE 2.59

DEFINITION Limit of a Function 1 2 Assume that f x exists for all x in some open interval containing a, except possibly at ➤ The value of d in the precise definition of f1x2 x a L a. We say that the limit of as approaches is , written a limit depends only on e. 1 2 = lim f x L, xSa ➤ Definitions of the one-sided limits 1 2 = 1 2 = if for any number e 7 0 there is a corresponding number d 7 0 such that lim f x L and lim f x L are xSa+ xSa- discussed in Exercises 39–43. 1 2 - 6 e 6 - 6 d f x L whenever 0 x a .

EXAMPLE 2 Finding D for a given E using a graphing utility Let 1 2 = 3 - 2 + - 1 2 = f x x 6x 12x 5 and demonstrate that lim f x 3 as follows. xS2 For the given values of e, use a graphing utility to find a value of d 7 0 such that 1 2 - 6 e 6 - 6 d f x 3 whenever 0 x 2 . e = e = 1 a. 1 b. 2 SOLUTION 1 2 - 6 e = 1 2 a. The condition f x 3 1 implies that f x lies between 2 and 4. Using a graphing utility, we graph f and the lines y = 2 and y = 4 (Figure 2.60). These lines intersect the graph of f at x = 1 and at x = 3. We now sketch the vertical lines

2.7 Precise Definitions of Limits 123

= = 1 2 x 1 and x 3 and observe that f x is within 1 unit of 3 whenever x is within 1 unit of 2 on the x-axis (Figure 2.60). Therefore, with e = 1, we can choose any d with 0 6 d … 1. 1 2 - 6 e = 1 1 2 b. The condition f x 3 2 implies that f x lies between 2.5 and 3.5 on the y-axis. We now find that the lines y = 2.5 and y = 3.5 intersect the graph of f at x ≈ 1.21 and x ≈ 2.79 (Figure 2.61). Observe that if x is less than 0.79 units from 1 2 2 on the x-axis, then f x is less than a half-unit from 3 on the y-axis. Therefore, with e = 1 d 6 d … 2 we can choose any with 0 0.79.

y y

y f (x) 5

y f (x) 4

3.5 f (x) 3 1 3 f (x) 3 q 3 2.5 2

0 x 1 2 3 0 1.212 2.79 4 x

0 x 2 1 0 x 2 0.79 FIGURE 2.60 FIGURE 2.61 This procedure could be repeated for smaller and smaller values of e 7 0. For each

value of e, there exists a corresponding value of d, proving that the limit exists. Related Exercises 13–14 ➤

6 - 6 d - d + d QUICK CHECK 2 For the function f The inequality 0 x a means that x lies between a and a with ≠ 1 - d + d2 a given in Example 2, estimate a value of x a. We say that the interval a , a is symmetric about because a is the d 7 1 2 - 6 midpoint of the interval. Symmetric intervals are convenient, but Example 3 demonstrates

0 satisfying f x 3 0.25 whenever 0 6 x - 2 6 d. ➤ that we don’t always get symmetric intervals without a bit of extra work.

y EXAMPLE 3 Finding a symmetric interval Figure 2.62 shows the graph of g with lim g1x2 = 3. For each value of e, find the corresponding values of d 7 0 that satisfy S 6 x 2 the condition

y g(x) g1x2 - 3 6 e whenever 0 6 x - 2 6 d. a. e = 2 3 b. e = 1 2 c. For any given value of e, make a conjecture about the corresponding values of d that satisfy the limit condition. 1 SOLUTION 0 x 12 6 a. With e = 2, we need a value of d 7 0 such that g1x2 is within 2 units of 3, which means d = FIGURE 2.62 between 1 and 5, whenever x is less than units from 2. The horizontal lines y 1 and y = 5 intersect the graph of g at x = 1 and x = 6. Therefore, g1x2 - 3 6 2 if x lies in the interval 11, 62 with x ≠ 2 (Figure 2.63a). However, we want x to lie in an interval that is symmetric about 2. We can guarantee that g1x2 - 3 6 2 only if x is less than 1 unit away from 2, on either side of 2 (Figure 2.63b). Therefore, with e = 2, we take d = 1 or any smaller positive number.

124 Chapter 2 r Limits

y y

5 5

y g(x) y g(x) g(x) 3 2 3 3

1 1

0 1 2 3 6 x 0 1 2 3 6 x

Values of x such that Symmetric interval 0 x 2 1 g(x) 3 2 that guarantees g(x) 3 2 (a) (b) FIGURE 2.63

b. With e = 1, g1x2 must lie between 2 and 4 (Figure 2.64a). This implies that x must be within a half-unit to the left of 2 and within 2 units to the right of 2. Therefore, g1x2 - 3 6 1 provided x lies in the interval 11.5, 42. To obtain a symmetric inter- d = 1 val about 2, we take 2 or any smaller positive number. Then we are guaranteed 1 2 - 6 6 - 6 1 that g x 3 1 when 0 x 2 2 (Figure 2.64b).

y y

4 4 y g(x) y g(x) g(x) 3 1 3 3

2 2

0 1.5 4 x 0 1.5 2.5 4 x

Values of x such that Symmetric interval 0 x 2 q g(x) 3 1 that guarantees g(x) 3 1 (a) (b) FIGURE 2.64 c. From parts (a) and (b), it appears that if we choose d … e>2, the limit condition is sat-

isfied for any e 7 0. Related Exercises 15–18 ➤

Limit Proofs 1 2 = We use the following two-step process to prove that lim f x L. xSa f1x2 = L Steps for proving that lim xSa ➤ D e 1 2 - 6 e The first step of the limit-proving process 1. Find . Let be an arbitrary positive number. Use the inequality f x L is the preliminary work of finding a to find a condition of the form x - a 6 d, where d depends only on the candidate for d. The second step verifies value of e. that the d found in the first step actually e 7 6 - 6 d works. 2. Write a proof. For any 0, assume 0 x a and use the rela- e d 1 2 - 6 e tionship between and found in Step 1 to prove that f x L .

2.7 Precise Definitions of Limits 125

EXAMPLE 4 Limit of a linear function Prove that lim 14x - 152 = 1 using the S precise definition of a limit. x 4 SOLUTION Step 1: Find d. In this case, a = 4 and L = 1. Assuming e 7 0 is given, we use 14x - 152 - 1 6 e to find an inequality of the form x - 4 6 d. If 14x - 152 - 1 6 e, then 4x - 16 6 e 4 x - 4 6 e Factor 4x - 16. e x - 4 6 . Divide by 4 and identify d = e>4. 4 We have shown that 14x - 152 - 1 6 e implies x - 4 6 e>4. Therefore, a plausible relationship between d and e is d = e>4. We now write the actual proof. Step 2: Write a proof. Let e 7 0 be given and assume 0 6 x - 4 6 d where d = e>4. The aim is to show that 14x - 152 - 1 6 e for all x such that 0 6 x - 4 6 d. We simplify 14x - 152 - 1 and isolate the x - 4 term: 14x - 152 - 1 = 4x - 16 = - 4 (+x )+4* less than d = e>4 e 6 4a b = e. 4 We have shown that for any e 7 0, 1 2 - = 1 - 2 - 6 e 6 - 6 d f x L 4x 15 1 whenever 0 x 4 ,

provided 0 6 d … e>4. Therefore, lim 14x - 152 = 1. S ➤ x 4 Related Exercises 19–24

Justifying Limit Laws The precise definition of a limit is used to prove the limit laws in Theorem 2.3. Essential in several of these proofs is the triangle inequality, which states that x + y … x + y , for all real numbers x and y.

1 2 1 2 EXAMPLE 5 Proof of Limit Law 1 Prove that if lim f x and lim g x exist, then xSa xSa 3 1 2 + 1 24 = 1 2 + 1 2 lim f x g x lim f x lim g x . xSa xSa xSa ➤ 1 2 Because lim f x exists, if there exists a e 7 1 2 = SOLUTION Assume that 0 is given. Let lim f x L, which implies that there xSa S d 7 e 7 d 7 x a 0 for any given 0, then there exists a 1 0 such that d 7 e also exists a 0 for any given 2. e 1 2 - 6 6 - 6 d f x L whenever 0 x a . 2 1 1 2 = d 7 Similarly, let lim g x M, which implies there exists a 2 0 such that xSa e g1x2 - M 6 whenever 0 6 x - a 6 d . 2 2

126 Chapter 2 r Limits

➤ d = 5d d 6 6 - 6 d d … d The minimum value of a and b is denoted Let min 1, 2 and suppose 0 x a . Because 1, it follows that 5 6 = 5 6 6 - 6 d 1 2 - 6 e> d … d min a, b . If x min a, b , then x is 0 x a 1 and f x L 2. Similarly, because 2, it follows that the smaller of a and b. If a = b, then 6 - 6 d 1 2 - 6 e> 0 x a 2 and g x M 2. Therefore, x equals the common value of a and b. 0 1 1 2 + 1 22 - 1 + 2 0 = 0 1 1 2 - 2 + 1 1 2 - 2 0 In either case, x … a and x … b. f x g x L M f x L g x M Rearrange terms. … 0 1 2 - 0 + 0 1 2 - 0 f x L g x M Triangle inequality. e e 6 + = e. 2 2 We have shown that given any e 7 0, if 0 6 x - a 6 d, then ➤ Proofs of other limit laws are outlined in 0 1 1 2 + 1 22 - 1 + 2 0 6 e 3 1 2 + 1 24 = f x g x L M , which implies that lim f x g x

xSa Exercises 25 and 26. + = 1 2 + 1 2 ➤ L M lim f x lim g x . Related Exercises 25–28 xSa xSa

➤ Notice that for infinite limits, N plays Infinite Limits e 1 2 = ∞ 1 2 the role that plays for regular limits. It In Section 2.4, we stated that lim f x if f x grows arbitrarily large as x approaches S sets a tolerance or bound for the function x a 1 2 a. More precisely, this means that for any positive number N (no matter how large), f x 1 2 values f x . is larger than N if x is sufficiently close to a but not equal to a.

DEFINITION Two-Sided Infinite Limit 1 2 = ∞ The infinite limit lim f x means that for any positive number N, there exists a xSa corresponding d 7 0 such that 1 2 7 6 - 6 d f x N whenever 0 x a .

1 2 = ∞ As shown in Figure 2.65, to prove that lim f x , we let N represent any positive xSa number. Then we find a value of d 7 0, depending only on N, such that 1 2 7 6 - 6 d f x N whenever 0 x a . This process is similar to the two-step process for finite limits.

➤ 1 2 = - ∞ Precise definitions for lim f x , xSa y 1 2 = - ∞ 1 2 = ∞ lim f x , lim f x , xSa+ xSa+ 1 2 = - ∞ 1 2 = ∞ lim f x , and lim f x xSa- xSa- are given in Exercises 45–49. f (x)

f (x) N

O a ␦ x aa ␦ x

0 x a

Values of x such that f (x) N FIGURE 2.65

2.7 Precise Definitions of Limits 127

f1x2 = H Steps for proving that lim xSa D 1 2 7 1. Find . Let N be an arbitrary positive number. Use the statement f x N to find an inequality of the form x - a 6 d, where d depends only on N. 2. Write a proof. For any N 7 0, assume 0 6 x - a 6 d and use the rela- d 1 2 7 tionship between N and found in Step 1 to prove that f x N.

1 2 = 1 EXAMPLE 6 An Infinite Limit Proof Let f x . Prove that 1x - 222 1 2 = ∞ lim f x . xS2

SOLUTION 1 Step 1: Find d 7 0. Assuming N 7 0, we use the inequality 7 N to find d, 1x - 222 where d depends only on N. Taking reciprocals of this inequality, it follows that 1 1x - 222 6 N 1 x - 2 6 . Take the square root of both sides. 1N ➤ 2 2 = Recall that x x . 1 1 The inequality x - 2 6 has the form x - 2 6 d if we let d = . 1N 1N We now write a proof based on this relationship between d and N. 1 Step 2: Write a proof. Suppose N 7 0 is given. Let d = and assume 1 1 N 06 x - 2 6 d = . Squaring both sides of the inequality 1N 1 x - 2 6 and taking reciprocals, we have 1N 1 1x - 222 6 Square both sides. N 1 QUICK CHECK 3 In Example 6, if N is 7 N. Take reciprocals of both sides. 1 - 22

increased by a factor of 100, how must x 2 d ➤ 1 change? We see that for any positive N, if 0 6 x - 2 6 d = , then 1N 1 2 = 1 7 1 = ∞ f x N. It follows that lim . Note that 1x - 222 xS2 1x - 222 1 because d = , d decreases as N increases.

1N Related Exercises 29–32 ➤

Limits at Infinity 1 2 = Precise definitions can also be written for the limits at infinity lim f x L and 1 2 = xS∞ lim f x L. For discussion and examples, see Exercises 50 and 51. xS-∞

128 Chapter 2 r Limits

SECTION 2.7 EXERCISES Review Questions 11. Determining values of D from a graph The function f in the fig- 1 2 ≠ 1 2 = d 7 ure satisfies lim f x 6. Determine the largest value of 0 1. Suppose x lies in the interval 1, 3 with x 2. Find the smallest S positive value of d such that the inequality 0 6 x - 2 6 d x 3 satisfying each statement. is true. 6 - 6 d 1 2 - 6 a. If 0 x 3 , then f x 6 3. 1 2 1 2 2. Suppose f x lies in the interval 2, 6 . What is the smallest value 6 - 6 d 1 2 - 6 b. If 0 x 3 , then f x 6 1. e 1 2 - 6 e of such that f x 4 ? y 3. Which one of the following intervals is not symmetric about x = 5? a. 11, 92 b. 14, 62 c. 13, 82 d. 14.5, 5.52 9 4. Does the set 5x: 0 6 x - a 6 d6 include the point x = a? y ϭ f (x) Explain. 6 1 2 = 5. State the precise definition of lim f x L. xSa 1 2 - 6 e 6. Interpret f x L in words. 1 2 - 6 6 6 7. Suppose f x 5 0.1 whenever 0 x 5. Find d 7 1 2 - 6 all values of 0 such that f x 5 0.1 whenever 0 6 x - 2 6 d. 1 1 2 = ∞ 8. Give the definition of lim f x and interpret it using xSa pictures. 0 13 6x

Basic Skills 12. Determining values of D from a graph The function f in the D 1 2 = 9. Determining values of from a graph The function f in the figure satisfies lim f x 5. Determine the largest value of 1 2 = xS4 figure satisfies lim f x 5. Determine the largest value d 7 xS2 0 satisfying each statement. of d 7 0 satisfying each statement. 6 - 6 d 1 2 - 6 a. If 0 x 4 , then f x 5 1. 6 - 6 d 6 - 6 d 1 2 - 6 a. If 0 x 2 , y b. If 0 x 4 , then f x 5 0.5. 1 2 - 6 then f x 5 2. 8 y b. If 0 6 x - 2 6 d, 1 2 - 6 8 then f x 5 1.

y ϭ f (x) 5

5 y ϭ f (x)

1 0 12 x

0 14 8x 10. Determining values of D from a graph The function f in the T D E 1 2 = 3 + 1 2 = figure satisfies lim f x 4. Determine the largest value 13. Finding for a given using a graph Let f x x 3 and xS2 1 2 = e note that lim f x 3. For each value of , use a graphing utility of d 7 0 satisfying each statement. xS0 d 7 1 2 - 6 e 6 - 6 d y to find a value of 0 such that f x 3 whenever a. If 0 x 2 , 6 - 6 d 1 2 - 6 0 x 0 . Sketch graphs illustrating your work. then f x 4 1. 8 e = e = b. If 0 6 x - 2 6 d, a. 1 b. 0.5 1 2 - 6 > then f x 4 1 2. T 14. Finding D for a given E using a graph Let 1 2 = 3 - 2 + + 1 2 = y ϭ f (x) g x 2x 12x 26x 4 and note that lim g x 24. xS2 For each value of e, use a graphing utility to find a value of d 7 0 4 such that g1x2 - 24 6 e whenever 0 6 x - 2 6 d. Sketch graphs illustrating your work. a. e = 1 b. e = 0.5 1 15. Finding a symmetric interval The function f in the figure 1 2 = e d 7 satisfies lim f x 3. For each value of , find a value of 0 0 12 4 x xS2 such that 1 2 - 6 e 6 - 6 d f x 3 whenever 0 x 2 . (2) e = e = 1 a. 1 b. 2 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 2014– Pilot Program

2.7 Precise Definitions of Limits 129

c. For any e 7 0, make a conjecture about the corresponding x 2 - 7x + 12 d 22. lim = -1 value of satisfying (2). xS3 x - 3 y 2 = 2 2 = 23. lim x 0 (Hint: Use the identity x x .) xS0 6 1 - 22 = 2 2 = 24. lim x 3 0 (Hint: Use the identity x x .) xS3 1 2 = 1 2 = y ϭ f (x) 25. Proof of Limit Law 2 Suppose lim f x L and lim g x M. xSa xSa 3 1 2 - 1 24 = - Prove that lim f x g x L M. 3 xSa 1 2 = 26. Proof of Limit Law 3 Suppose lim f x L. Prove that xSa 3 1 24 = 1 lim cf x cL, where c is a constant. xSa f1x2 = x 27. Limit of a constant function and Give proofs of the 0 x 12 6 following theorems. = a. lim c c for any constant c 16. Finding a symmetric interval The function f in the figure satis- xSa 1 2 = e d 7 = fies lim f x 5. For each value of , find a value of 0 b. lim x a for any constant a xS4 xSa such that 1 2 = 28. Continuity of linear functions Prove Theorem 2.2: If f x + 1 2 = + 1 2 - 6 e 6 - 6 d f x 5 whenever 0 x 4 . (3) mx b, then lim f x ma b for constants m and b. (Hint: xSa e 7 d = e> a. e = 2 b. e = 1 For a given 0, let m .) Explain why this result implies c. For any e 7 0, make a conjecture about the corresponding that linear functions are continuous. d value of satisfying (3). 29–32. Limit proofs for infinite limits Use the precise definition of y infinite limits to prove the following limits. 7 1 1 29. lim = ∞ 30. lim = ∞ 6 xS4 1x - 422 xS-1 1x + 124 5 a 1 + b = ∞ a 1 - b = ∞ 4 31. lim 1 32. lim sin x xS0 x 2 xS0 x 4 3 2 Further Explorations 1 33. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. 0 1 2 3 4 5 6 7 x 1 2 = Assume a and L are finite numbers and assume lim f x L. xSa 2x 2 - 2 a. For a given e 7 0, there is one value of d 7 0 for which T 1 2 = 17. Finding a symmetric interval Let f x and note 1 2 - 6 e 6 - 6 d x - 1 f x L whenever 0 x a . 1 2 = d 7 1 2 = e b. The limit lim f x L means that given an arbitrary 0, that lim f x 4. For each value of , use a graphing utility S xSa x 1 e 7 1 2 - 6 e d 7 1 2 - 6 e we can always find an 0 such that f x L when- to find a value of 0 such that f x 4 whenever 6 - 6 d 0 6 x - 1 6 d. ever 0 x a . 1 2 = e 7 c. The limit lim f x L means that for any arbitrary 0, a. e = 2 b. e = 1 xSa d 7 1 2 - 6 e c. For any e 7 0, make a conjecture about the value of d that we can always find a 0 such that f x L when- satisfies the preceding inequality. ever 0 6 x - a 6 d. d. If x - a 6 d, then a - d 6 x 6 a + d. 1 x + 1if x … 3 T 1 2 = b 3 18. Finding a symmetric interval Let f x 1 1 D 1 2 = 2 - + + 7 2 x 2 if x 3 34. Finding algebraically Let f x x 2x 3. 1 2 = e and note that lim f x 2. For each value of , use a graphing a. For e = 0.25, find a corresponding value of d 7 0 satisfying xS3 d 7 1 2 - 6 e the statement utility to find a value of 0 such that f x 2 whenever 0 6 x - 3 6 d. 1 2 - 6 e 6 - 6 d f x 2 whenever 0 x 1 . e = 1 e = 1 a. 2 b. 4 1 2 = e 7 e 7 d b. Verify that lim f x 2 as follows. For any 0, find a c. For any 0, make a conjecture about the value of that xS1 satisfies the preceding inequality. corresponding value of d 7 0 satisfying the statement 1 2 - 6 e 6 - 6 d 19–24. Limit proofs Use the precise definition of a limit to prove the f x 2 whenever 0 x 1 . following limits. 35–38. Challenging limit proofs Use the definition of a limit to prove 1 + 2 = 1- + 2 = 19. lim 8x 5 13 20. lim 2x 8 2 the following results. xS1 xS3 1 1 x 2 - 16 35. lim = (Hint: As x S 3, eventually the distance between x 21. lim = 8 (Hint: Factor and simplify.) xS3 x 3 xS4 x - 4 and 3 will be less than 1. Start by assuming x - 3 6 1 and 1 1 show 6 .2 x 2 Copyright © 2014 Pearson Education, Inc. For Use Only in 2013 2014– Pilot Program

130 Chapter 2 r Limits

x - 4 Additional Exercises 36. lim = 4 (Hint: Multiply the numerator and xS4 1x - 2 44. The relationship between one-sided and two-sided limits denominator by 1x + 2.) Prove the following statements to establish the fact that 1 2 = 1 2 = 1 2 = lim f x L if and only if lim f x L and lim f x L. 1 xSa xSa- xSa+ 37. lim = 10 (Hint: To find d, you will need to bound x away 1 2 = 1 2 = 1 2 = xS1>10 a. If lim- f x L and lim+ f x L, then lim f x L. x xSa xSa xSa 1 2 = 1 2 = 1 2 = 1 1 b. If lim f x L, then lim f x L and lim f x L. from 0. So let ` x - ` 6 .) xSa xSa- xSa+ 10 20 45. Definition of one-sided infinite limits We say that 1 2 = - ∞ 1 1 lim f x if for any negative number N, there exists 38. lim = S + S 2 x a x 5 x 25 d 7 0 such that 39–43. Precise definitions for left- and right-sided limits 1 2 6 6 6 + d f x N whenever a x a . Use the following definitions. 7 1 2 = ∞ Assume f exists for all x near a with x a. We say that the a. Write an analogous formal definition for lim+ f x . limit of f 1x2 as x approaches a from the right of a is L xSa and write 1 2 = - ∞ b. Write an analogous formal definition for lim f x . 1 2 = e 7 d 7 S - lim f x L, if for any 0 there exists 0 such that x a S + 1 2 = ∞ x a c. Write an analogous formal definition for lim f x . 1 2 - 6 e 6 - 6 d xSa- f x L whenever 0 x a . 46–47. One-sided infinite limits Use the definitions given in Exercise 45 Assume f exists for all values of x near a with x 6 a. We say that to prove the following infinite limits. the limit of f 1x2 as x approaches a from the left of a is L and write 1 2 = e 7 d 7 lim f x L, if for any 0 there exists 0 such that 1 1 xSa- 46. lim = - ∞ 47. lim = ∞ S + - S - - 1 2 - 6 e 6 - 6 d x 1 1 x x 1 1 x f x L whenever 0 a x . 1 2 = - ∞ 48–49. Definition of an infinite limit We write lim f x if for 39. Comparing definitions Why is the last inequality in the defini- xSa 1 2 = 6 - 6 d d 7 tion of lim f x L, namely, 0 x a , replaced any negative number M there exists a 0 such that xSa 6 - 6 d 1 2 = 1 2 6 6 - 6 d with 0 x a in the definition of lim+ f x L? f x M whenever 0 x a . xSa 40. Comparing definitions Why is the last inequality in the defini- Use this definition to prove the following statements. 1 2 = 6 - 6 d tion of lim f x L, namely, 0 x a , replaced with xSa - - 6 - 6 d 1 2 = 2 10 = - ∞ = - ∞ 0a x in the definition of lim- f x L? 48. lim 49. lim xSa xS1 1x - 122 xS-2 1x + 224 41. One-sided limit proofs Prove the following limits for 50–51. Definition of a limit at infinity The limit at infinity - 6 1 2 = e 7 7 3x 4if x 0 lim f x L means that for any 0, there exists N 0 such that 1 2 = b S f x - Ú x ∞ 2x 4if x 0. 1 2 - 6 e 7 f x L whenever x N. 1 2 = - 1 2 = - a. lim f x 4 b. lim f x 4 xS0+ xS0- Use this definition to prove the following statements. 1 2 = - c. lim f x 4 xS0 10 50. lim = 0 42. Determining values of D from a graph The function f in the xS∞ x 1 2 = 1 2 = figure satisfies lim f x 0 and lim f x 1. Determine a xS2+ xS2- 2x + 1 d 7 51. lim = 2 value of 0 satisfying each statement. xS∞ x 1 2 - 6 6 - 6 d a. f x 0 2 whenever 0 x 2 1 2 - 6 6 - 6 d 52–53. Definition of infinite limits at infinity We say that b. f x 0 1 whenever 0 x 2 1 2 = ∞ 1 2 - 6 6 - 6 d lim f x if for any positive number M, there is a c. f x 1 2 whenever 0 2 x xS∞ 1 2 - 6 6 - 6 d 7 d. f x 1 1 whenever 0 2 x corresponding N 0 such that 1 2 7 7 y f x M whenever x N.

6 Use this definition to prove the following statements. x 52. lim = ∞ xS∞ 100 y ϭ f (x) x 2 + x 53. lim = ∞ xS∞ x 1 54. Proof of the Squeeze Theorem Assume the functions f, g, and h 1 2 … 1 2 … 1 2 0 x satisfy the inequality f x g x h x for all values of x near a, 12 6 1 2 = 1 2 = except possibly at a. Prove that if lim f x lim h x L, then S S lim g1x2 = L. x a x a 43. One-sided limit proof Prove that lim 1x = 0. S xS0+ x a

Review Exercises 131

55. Limit proof Suppose f is defined for all values of x near a, x except possibly at a. Assume for any integer N 7 0, there is 57. Prove that lim does not exist. xS0 x 7 1 2 - 6 > another integer M 0 such that f x L 1 N whenever 58. Let - 6 > 1 2 = x a 1 M. Prove that lim f x L using the precise xSa definition of a limit. 1 2 = e 0if x is rational f x f 1x2 3 L 1if x is irrational. 56–58. Proving that xlimSa Use the following definition 1 2 Prove that lim f x does not exist for any value of a. (Hint: for the nonexistence of a limit. Assume f is defined for all values of x xSa 1 2 ≠ 1 1 2 = e = near a, except possibly at a. We say that lim f x L if for some Assume lim f x L for some values of a and L and let .) xSa S 2 e 7 d 7 x a 0 there is no value of 0 satisfying the condition 59. A continuity proof Suppose f is continuous at a and assume 1 2 7 d 7 1 2 - 6 e 6 - 6 d f x L whenever 0 x a . f a 0. Show that there is a positive number 0 for which 1 2 7 1 - d + d2 f x 0 for all values of x in a , a . (In other words, f 1 2 ≠ 56. For the following function, note that lim f x 3. Find a value xS2 is positive for all values of x in the domain sufficiently close to a.) of e 7 0 for which the preceding condition for nonexistence is satisfied. QUICK CHECK ANSWERS d = 1 d = d

y 1. 50 or smaller 2. 0.62 or smaller 3. must decrease by a factor of 1100 = 10 (at least). ➤ 6

4 y ϭ f (x) 3

0 1234 x

CHAPTER 2 REVIEW EXERCISES

1. Explain why or why not Determine whether the following state- 2. Estimating limits graphically Use the graph of f in the figure to ments are true and give an explanation or counterexample. find the following values, if possible. x - 1 1- 2 1 2 1 2 1 2 a. f 1 b. lim f x c. lim f x d. lim f x a. The rational function has vertical asymptotes at S- - S- + S- 2 - x 1 x 1 x 1 x 1 1 2 1 2 1 2 1 2 e. f 1 f. lim f x g. lim f x h. lim f x x = -1 and x = 1. xS1 xS2 xS3- 1 2 1 2 b. Numerical or graphical methods always produce good i. lim f x j. lim f x 1 2 xS3+ xS3 estimates of lim f x . xSa 1 2 1 2 y c. The value of lim f x , if it exists, is found by calculating f a . xSa 1 2 = ∞ 1 2 = - ∞ 1 2 d. If lim f x or lim f x , then lim f x does S S S 6 x a x a x a ϭ not exist. y f (x) 1 2 1 2 = ∞ 5 e. If lim f x does not exist, then either lim f x or xSa xSa 1 2 = - ∞ lim f x . 4 xSa f. If a function is continuous on the intervals 1a, b2 and 1b, c2, where 3 a 6 b 6 c, then the function is also continuous on 1a, c2. 1 2 2 g. If lim f x can be calculated by direct substitution, then f is xSa continuous at x = a. 1

Ϫ1 1234 x